1. hongthamvp@gmail.com Su tm v bin son MC LC Chiu lch ca tia sng..................................................................................................................................222 TNG HP BI TP VT L 11 CHNG I:IN TCH.IN TRNG CH 1:LC TNG TC TNH IN DNG 1: TNG TC GIA HAI IN TCH IM NG YN A.L THUYT 1.Lc tng tc gia 2 in tch im. Lc tng tc gia hai in tch im ql v q2 (nm yn, t trong chn khng) cch nhau on r c: phng l ng thng ni hai in tch. chiu l: chiu lc y nu qlq2 > 0 (cng du). chiu lc ht nu qlq2 < 0 (tri du). ln: * t l thun vi tch cc ln ca hai in tch, * t l nghch vi bnh phng khong cch gia chng. Trong : k = 9.109 N.m2 /C2 . q 1 , q 2 : ln hai in tch (C ) r: khong cch hai in tch (m) 1 F= 1 2 2 q q k r

2. hongthamvp@gmail.com Su tm v bin son : hng s in mi . Trong chn khng v khng kh =1 Ch : a) in tch im : l vt m kch thc cc vt cha in tch rt nh so vi khong cch gia chng. -Cng thc trn cn p dng c cho trng hp cc qu cu ng cht , khi ta coi r l khong cch gia tm hai qu cu. 2. in tch q ca mt vt tch in: e.nq = + Vt thiu electron (tch in dng): q = + n.e + Vt tha electron (tch in m): q = n.e Vi: C10.6,1e 19 = : l in tch nguyn t. n : s ht electron b tha hoc thiu. 3.Mt s hin tng Khi cho 2 qu cu nh nhim in tip xc sau tch nhau ra th tng in tch chia u cho mi qu cu Hin tng xy ra tng t khi ni hai qu cu bng dy dn mnh ri ct b dy ni Khi chm tay vo qu cu nh dn in tch in th qu cu mt in tch v tr v trung ha B.BI TP T LUN Bi 1. Hai in tch C10.2q 8 1 = , C10q 8 2 = t cch nhau 20cm trong khng kh. Xc nh ln v v hnh lc tng tc gia chng? S: N10.5,4 5 Bi 2. Hai in tch C10.2q 6 1 = , C10.2q 6 2 = t ti hai im A v B trong khng kh. Lc tng tc gia chng l 0,4N. Xc nh khong cch AB, v hnh lc tng tc . S: 30cm Bi 3. Hai in tch t cch nhau mt khong r trong khng kh th lc tng tc gia chng l 3 10.2 N. Nu vi khong cch m t trong in mi th lc tng tc gia chng l 3 10 N. a/ Xc nh hng s in mi ca in mi. b/ lc tng tc gia hai in tch khi t trong in mi bng lc tng tc khi t trong khng kh th phi t hai in tch cch nhau bao nhiu? Bit trong khng kh hai in tch cch nhau 20cm. S: 2= ; 14,14cm. Bi 4. Trong nguyn t hir (e) chuyn ng trn u quanh ht nhn theo qu o trn c bn knh 5.10 -9 cm. a. Xc nh lc ht tnh in gia (e) v ht nhn. b. Xc nh tn s ca (e) S: F=9.10-8 N b.0,7.1016 Hz Bi 5. Mt qu cu c khi lng ring (aKLR) = 9,8.103 kg/m3 ,bn knh R=1cm tch in q = -10 -6 C c treo vo u mt si dy mnh c chiu di l =10cm. Ti im treo c t mt in tch m q0 = - 10 -6 C .Tt c t trong du c KLR D= 0,8 .103 kg/m3 ,hng s in mi =3.Tnh lc cng ca dy? Ly g=10m/s2 . S:0,614N Bi 6. Hai qu cu nh, ging nhau, bng kim loi. Qu cu A mang in tch 4,50 C; qu cu B mang in tch 2,40 C. Cho chng tip xc nhau ri a chng ra cch nhau 1,56 cm. Tnh lc tng tc in gia chng. ____________________________________________________________________________________________ DNG 2: LN IN TCH. A.L THUYT 2 3. hongthamvp@gmail.com Su tm v bin son Dng 2: Xc nh ln v du cc in tch. - Khi gii dng BT ny cn ch : Hai in tch c ln bng nhau th: 21 qq = Hai in tch c ln bng nhau nhng tri du th: 21 qq = Hai in tch bng nhau th: 21 qq = . Hai in tch cng du: 212121 q.qq.q0q.q => . Hai in tch tri du: 212121 q.qq.q0q.q =< - p dng h thc ca nh lut Coulomb tm ra 21 q.q sau ty iu kin bi ton chng ra s tm c q1 v q2. - Nu bi ch yu cu tm ln th ch cn tm 21 q;q 2.1/Bi tp v d: Hai qu cu nh tch in c ln bng nhau, t cch nhau 5cm trong chn khng th ht nhau bng mt lc 0,9N. Xc nh in tch ca hai qu cu . Tm tt: 21 qq = m05,0cm5r == N9,0F = , lc ht. ?q?q 21 == Gii. Theo nh lut Coulomb: 2 21 r q.q .kF = k r.F q.q 2 21 = 14 9 2 21 10.25 10.9 05,0.9,0 q.q == M 21 qq = nn 142 1 10.25q = C10.5qq 7 12 == Do hai in tch ht nhau nn: C10.5q 7 1 = ; C10.5q 7 2 = hoc: C10.5q 7 1 = ; C10.5q 7 2 = B.BI TP T LUN Bi 1. Hai in tch im bng nhau, t trong chn khng, cch nhau 10 cm. Lc y gia chng l 9.10-5 N. a/ Xc nh du v ln hai in tch . b/ lc tng cc gia hai in tch tng 3 ln th phi tng hay gim khong cch gia hai in tch bao nhiu ln? V sao? Xc nh khong cch gia hai in tch lc . S: a/ C10qq 8 21 == ; hoc C10qq 8 21 == b/Gim 3 ln; cm77,5'r Bi 2. Hai in tch c ln bng nhau, t cch nhau 25cm trong in mi c hng s in mi bng 2 th lc tng tc gia chng l 6,48.10-3 N. a/ Xc nh ln cc in tch. b/ Nu a hai in tch ra khng kh v vn gi khong cch th lc tng tc gia chng thay i nh th no? V sao? c/ lc tng tc ca hai in tch trong khng kh vn l 6,48.10-3 N th phi t chng cch nhau bng bao nhiu? 3 4. hongthamvp@gmail.com Su tm v bin son S: a/ C10.3qq 7 21 == ; b/ tng 2 ln c/ cm36,35.rr mkk = . Bi 3. Hai vt nh tch in t cch nhau 50cm, ht nhau bng mt lc 0,18N. in tch tng cng ca hai vt l 4.10-6 C. Tnh in tch mi vt? S: = = =+ = =+ = C10.5q C10q 10.4qq 10.5q.q 10.4qq 10.5q.q 6 2 6 1 6 21 12 21 6 21 12 21 Bi 5. Hai in tch im c ln bng nhau t trong chn khng, cch nhau 1 khong 5 cm, gia chng xut hin lc y F = 1,6.10-4 N. a.Hy xc nh ln ca 2 in tch im trn? b. lc tng tc gia chng l 2,5.10-4 N th khong cch gia chng l bao nhiu? S: 667nC v 0,0399m Bi 6 Hai vt nh t trong khng kh cch nhau mt on 1m, y nhau mt lc F= 1,8 N. in tch tng cng ca hai vt l 3.10-5 C. Tm in tch ca mi vt. S: 5 1 2.10q C = ; 5 2 10q C = Bi 7. Hai qu cu kim loi nh nh nhau mang cc in tch q1 v q2 t trong khng kh cch nhau 2 cm, y nhau bng mt lc 2,7.10-4 N. Cho hai qu cu tip xc nhau ri li a v v tr c, ch y nhau bng mt lc 3,6.10-4 N. Tnh q1, q2 ? S: 9 1 2.10q C = ; 9 2 6.10q C = v 9 1 2.10q C = ; 9 2 6.10q C = v o li Bi 8. Hai qu cu nh ging nhau bng kim loi c khi lng 50g c treo vo cng mt im bng 2 si ch nh khng gin di 10cm. Hai qu cu tip xc nhau tch in cho mt qu cu th thy hai qu cu y nhau cho n khi 2 dy treo hp vi nhau mt gc 600 .Tnh in tch m ta truyn cho cc qu cu qu cu.Cho g=10 m/s2 . S: q=3,33C Bi 9. Mt qu cu nh c m = 60g ,in tch q = 2. 10 -7 C c treo bng si t mnh. pha di n 10 cm cnt mt in tch q2 nh th no sc cng ca si dy tng gp i? S: q=3,33C Bi 10. Hai qu cu nh tch in q1= 1,3.10 -9 C ,q2 = 6,5.10-9 C t cch nhau mt khong r trong chn khng th y nhau vi mt nhng lc bng F. Cho 2 qu cu y tip xc nhau ri t cch nhau cng mt khong r trong mt cht in mi th lc y gia chng vn l F. a, Xc nh hng s in mi ca cht in mi . b, Bit F = 4,5.10 -6 N ,tm r S: =1,8. r=1,3cm ----------------------------------------------------------------------------------------------------------------------------- 4 5. hongthamvp@gmail.com Su tm v bin son DNG 3: TNG TC CA NHIU IN TCH A.L THUYT Dng 3: Hp lc do nhiu in tch tc dng ln mt in tch. * Phng php: Cc bc tm hp lc oF do cc in tch q1; q2; ... tc dng ln in tch qo: Bc 1: Xc nh v tr im t cc in tch (v hnh). Bc 2: Tnh ln cc lc ...F;F 2010 , Fno ln lt do q1 v q2 tc dng ln qo. Bc 3: V hnh cc vect lc 2010 F;F .... 0nF uuuv Bc 4: T hnh v xc nh phng, chiu, ln ca hp lc oF . + Cc trng hp c bit: 2 Lc: Gc bt k: l gc hp bi hai vect lc. 2 2 2 0 10 20 10 202 .cosF F F F F = + + 3.1/ Bi tp v d: Trong chn khng, cho hai in tch C10qq 7 21 == t ti hai im A v B cch nhau 8cm. Ti im C nm trn ng trung trc ca AB v cch AB 3cm ngi ta t in tch C10q 7 o = . Xc nh lc in tng hp tc dng ln qo. Tm tt: 5 6. hongthamvp@gmail.com Su tm v bin son C10q C10q 7 2 7 1 = = cm3AH;cm8AB;C10q 7 o === ?Fo = Gii: V tr cc in tch nh hnh v. + Lc do q1 tc dng ln qo: N036,0 05,0 10.10 10.9 AC qq kF 2 77 9 2 01 10 === + Lc do q2 tc dng ln qo: N036,0FF 1020 == ( do 21 qq = ) + Do 1020 FF = nn hp lc Fo tc dng ln qo: N10.6,57 5 4 .036,0.2F AC AH .F.2Acos.F.2Ccos.F2F 3 o 1010110o == === + Vy oF c phng // AB, cng chiu vi vect AB (hnh v) v c ln: N10.6,57F 3 o = B.BI TP T LUN Bi 1. Cho hai in tch im 7 7 1 22.10 ; 3.10q C q C = = t ti hai im A v B trong chn khng cch nhau 5cm. Xc nh lc in tng hp tc dng ln 7 2.10oq C = trong hai trng hp: a/ oq t ti C, vi CA = 2cm; CB = 3cm. b/ oq t ti D vi DA = 2cm; DB = 7cm. S: a/ oF 1,5N= ; b/ 0,79F N= . Bi 2. Hai in tch im 8 8 1 23.10 ; 2.10q C q C = = t ti hai im A v B trong chn khng, AB = 5cm. in tch 8 2.10oq C = t ti M, MA = 4cm, MB = 3cm. Xc nh lc in tng hp tc dng ln oq . S: 3 oF 5,23.10 N . 6 7. hongthamvp@gmail.com Su tm v bin son Bi 3. Trong chn khng, cho hai in tch 7 1 2 10q q C = = t ti hai im A v B cch nhau 10cm. Ti im C nm trn ng trung trc ca AB v cch AB 5cm ngi ta t in tch C10q 7 o = . Xc nh lc in tng hp tc dng ln qo. S: 0,051oF N . Bi 4. C 3 din tch im q1 =q2 = q3 =q = 1,6.10-6 c t trong chn khng ti 3 nh ca mt tam gic u ABC cnh a= 16 cm.Xc nh lc in tng hp tc dng ln mi in tch. Bi 5. Ba qu cu nh mang in tch q1 = 6.10 -7 C,q2 = 2.10 -7 C,q3 = 10 -6 C theo th t trn mt ng thng nhng trong nc nguyn cht c = 81..Khong cch gia chng l r12 = 40cm,r23 = 60cm.Xc nh lc in tng hp tc dng ln mi qu cu. Bi 6. Ba in tch im q1 = 4. 10-8 C, q2 = -4. 10-8 C, q3 = 5. 10-8 C. t trong khng kh ti ba nh ca mt tam gic u cnh 2 cm. Xc nh vect lc tc dng ln q3 ? Bi 7. Hai in tch q1 = 8.10-8 C, q2 = -8.10-8 C t ti A v B trong khng kh (AB = 10 cm). Xc nh lc tc dng ln q3 = 8.10-8 C , nu: a. CA = 4 cm, CB = 6 cm. b. CA = 14 cm, CB = 4 cm. c. CA = CB = 10 cm.d. CA=8cm, CB=6cm. Bi 8. Ngi ta t 3 in tch q1 = 8.10-9 C, q2 = q3 = -8.10-9 C ti ba nh ca mt tam gic u cnh 6 cm trong khng kh. Xc nh lc tc dng ln in tch q0 = 6.10-9 C t tm O ca tam gic. S:7,2.10-5 N ___________________________________________________________________________________________ DNG 4: CN BNG CA IN TCH A.L THUYT Dng 4: in tch cn bng. 7 8. hongthamvp@gmail.com Su tm v bin son * Phng php: Hai in tch: Hai in tch 1 2;q q t ti hai im A v B, hy xc nh im C t in tch oq oq cn bng: - iu kin cn bng ca in tch oq : 10 20 0oF F F= + = 10 20F F= = 2010 2010 FF FF )2( )1( + Trng hp 1: 1 2;q q cng du: T (1) C thuc on thng AB: AC + BC = AB (*) Ta c: 1 2 2 2 1 2 q q r r = + Trng hp 2: 1 2;q q tri du: T (1) C thuc ng thng AB: AC BC AB = (* ) Ta cng vn c: 1 2 2 2 1 2 q q r r = - T (2) 2 2 2 1. . 0q AC q BC = (**) - Gii h hai pt (*) v (**) hoc (* ) v (**) tm AC v BC. * Nhn xt: - Biu thc (**) khng cha oq nn v tr ca im C cn xc nh khng ph thuc vo du v ln ca oq . A BC r1 r2 q1 q2 q0 A B C r1 r2 q0 q1 q2 8 9. hongthamvp@gmail.com Su tm v bin son -V tr cn bng nu hai in tch tri du th im cn bng nm ngoi on AB v pha in tch c ln nh hn.cn nu hai in tch cng du th nm gia on ni hai in tch. Ba in tch: - iu kin cn bng ca q0 khi chu tc dng bi q1, q2, q3: + Gi 0F l tng hp lc do q1, q2, q3 tc dng ln q0: 03020100 =++= FFFF + Do q0 cn bng: 00 =F = =+ += =++ 30 30 30 2010 302010 0 0 FF FF FF FFF FFF B.BI TP T LUN Bi 1. Hai in tch 8 8 1 22.10 ; 8.10q C q C = = t ti A v B trong khng kh, AB = 8cm. Mt in tch oq t ti C. Hi: a/ C u oq cn bng? b/ Du v ln ca oq 1 2;q q cng cn bng? S: a/ CA = 8cm; CB = 16cm; b/ 8 8.10oq C = . Bi 2. Hai in tch 8 7 1 22.10 ; 1,8.10q C q C = = t ti A v B trong khng kh, AB = 8cm. Mt in tch 3q t ti C. Hi: a/ C u 3q cn bng? b*/ Du v ln ca 3q 1 2;q q cng cn bng? S: a/ CA = 4cm; CB = 12cm; b/ 8 3 4,5.10q C = . Bi 3*. Hai qu cu nh ging nhau, mi qu c in tch q v khi lng m = 10g c treo bi hai si dy cng chiu di 30l cm= vo cng mt im O. Gi qu cu 1 c nh theo phng thng ng, dy treo qu cu 2 s b lch gc 60o = so vi phng thng ng. Cho 2 10 /g m s= . Tm q? S: 6 10 mg q l C k = = Bi 4. Hai in tch im q1 = 10-8 C, q2 = 4. 10-8 C t ti A v B cch nhau 9 cm trong chn khng. a. Xc nh lc tng tc gia hai in tch? b. Xc nh vecto lc tc dng ln in tch q0 = 3. 10-6 C t ti trung im AB. c. Phi t in tch q3 = 2. 10-6 C ti u in tch q3 nm cn bng? Bi 5. Hai in tch im q1 = q2 = -4. 10-6 C, t ti A v B cch nhau 10 cm trong khng kh. Phi t in tch q3 = 4. 10-8 C ti u q3 nm cn bng? 9 10. hongthamvp@gmail.com Su tm v bin son Bi 6. Hai in tch q1 = - 2. 10-8 C, q2= -8. 10-8 C t ti A v B trong khng kh, AB = 8 cm.Mt in tch q3 t ti C. Hi: a. C u q3 cn bng? b. Du v ln ca q3 q1 v q2 cng cn bng ? Bi 7: Ba qu cu nh khi lng bng nhau v bng m, c treo vo 3 si dy cng chiu di l v c buc vo cng mt im. Khi c tch mt in tch q nh nhau, chng y nhau v xp thnh mt tam gic u c cnh a. Tnh in tch q ca mi qu cu? S: 3 2 2 3(3 ) ma g k l a Bi 8:Cho 3 qu cu ging ht nhau, cng khi lng m v in tch. trng thi cn bng v tr ba qu cu v im treo chung O to thnh t din u. Xc nh in tch mi qu cu? S: 6 mg q l k = CH 2:BI TP V IN TRNG DNG I:IN TRNG DO MT IN TCH IM GY RA A.L THUYT * Phng php: -Nm r cc yu t ca Vct cng in trng do mt in tch im q gy ra ti mt im cch in tch khong r: E : + im t: ti im ta xt + phng: l ng thng ni im ta xt vi in tch + Chiu: ra xa in tch nu q > 0, hng vo nu q < 0 + ln: 2 r q kE = - Lc in trng: EF q= , ln EqF = Nu q > 0 th EF ; Nu q < 0 th EF Ch : Kt qu trn vn ng vi in trng mt im bn ngoi hnh cu tch in q, khi ta coi q l mt in tch im t ti tm cu. Bi 1. Mt in tch im q = 10-6 C t trong khng kh a. Xc nh cng in trng ti im cch in tch 30cm, v vect cng in trng ti im ny b. t in tch trong cht lng c hng s in mi = 16. im c cng in trng nh cu a cch in tch bao nhiu. Bi 2: Cho hai im A v B cng nm trn mt ng sc ca in trng do mt in tch im q > 0 gy ra. Bit ln ca cng in trng ti A l 36V/m, ti B l 9V/m. a. Xc nh cng in trng ti trung im M ca AB. b. Nu t ti M mt in tch im q0 = -10-2 C th lnn lc in tc dng ln q0 l bao nhiu? Xc nh phng chiu ca lc. 10 11. hongthamvp@gmail.com Su tm v bin son q A M B EM Hng dn gii: Ta c: A 2 q E k 36V / m OA = = (1) B 2 q E k 9V / m OB = = (2) M 2 q E k OM = (3) Ly (1) chia (2) 2 OB 4 OB 2OA OA = = . Ly (3) chia (1) 2 M A E OA E OM = Vi: OA OB OM 1,5OA 2 + = = 2 M M A E OA 1 E 16V E OM 2,25 = = = b. Lc t tc dng ln qo: M0F q E= u v q0 < u u c. Khi 1 2E E u u 2 2 1 2E E E= + E u hp vi 1E u mt gc xc nh bi: 2 1 E tan E = d. Khi E1 = E2 v 1 2E ,E = u 1E 2E cos 2 = E u hp vi 1E u mt gc 2 e.Trng hp gc bt k p dng nh l hm cosin. - Nu bi i hi xc nh lc in trng tc dng ln in tch th p dng cng thc: Eq=F Bi 1: Cho hai in tch q1 = 4.10 -10 C, q2 = -4.10 -10 C t A,B trong khng kh, AB = a = 2cm. Xc nh vc t cng in trng ti: a) H l trungim ca AB. b) M cch A 1cm, cch B 3cm. c) N hp vi A,B thnh tam gic u. S: a.72.103 (V/m); b.32. 103 (V/m); c.9000(V/m); Bi 2: Hai in tch q1=8.10 -8 C, q2= -8.10 -8 C t ti A, B trong khng kh., AB=4cm. Tm vct cng in trng ti C vi: a) CA = CB = 2cm. b) CA = 8cm; CB = 4cm. 12 13. hongthamvp@gmail.com Su tm v bin son c) C trn trung trc AB, cch AB 2cm, suy ra lc tc dng ln q=2.10 -9 C t ti C. S: E song song vi AB, hng t A ti B c ln E=12,7.105V/m; F=25,4.10-4N) 13 14. hongthamvp@gmail.com Su tm v bin son Bai 3: Hai in tch +q v q (q >0) t ti hai im A v B vi AB = 2a. M l mt im nm trn ng trung trc ca AB cch AB mt on x. a. Xc nh vect cng in trng ti M b. Xc nh x cng in trng ti M cc i, tnh gi tr Hng dn gii: E1 M E E2 x a a A H B a. Cng in trng ti M: 1E E E2= + ta c: q E E k1 2 2 2 a x = = + Hnh bnh hnh xc nh E l hnh thoi: E = 2E1cos ( ) 2kqa 3/2 a x = + (1) b. T (1) Thy Emax th x = 0: Emax = 2kq E1 2 2 a x = + b) Lc cng dy: mg 2 T R 2.10 N cos = = = Bi 4 Hai in tch q1 = q2 = q >0 t ti A v B trong khng kh. cho bit AB = 2a E u 2E u 1E u M h q1 a a q2 A H B a) Xc nh cng in trng ti im M trn ng trung trc ca AB cch Ab mt on h. b) nh h EM cc i. Tnh gi tr cc i ny. Hng dn gii: a) Cng in trng ti M: E E E1 2= + Ta c: q E E k1 2 2 2 a x = = + 14 15. hongthamvp@gmail.com Su tm v bin son Hnh bnh hnh xc nh E l hnh thoi: E = 2E1cos ( ) 2kqh 3/2 2 2 a h = + b) nh h EM t cc i: ( ) ( ) 2 2 4 2 a a a .h2 2 2 3a h h 3. 2 2 4 3 3/227 3 32 2 4 2 2 2 2 a h a h a h a h 4 2 + = + + + + Do : 2kqh 4kq EM 23 3 3 3a2 a h 2 = EM t cc i khi: ( ) 2 a a 4kq2 h h EM 2max2 2 3 3a = = = Bi 5 Ti 3 nh ABC ca t din u SABC cnh a trong chn khng c ba in ch im q ging nhau (q0 t A, C, hai in tch q3=q4=-q t B v D. Tnh ln cng in trng ti tm O ca hnh lp phng. (S: 2 16 3 3 kq a ) DNG 3: CNG IN TRNG TNG HP TRIT TIU Tng qut: E=E1+E2+.......................+En= 0 Trng hp ch c haiin tch gy in trng: 1/ Tm v tr cng in trng tng hp trit tiu: a/ Trng hp 2 in tch cng du:( q 1 ,q 2 > 0 ) : q 1 t ti A, q 2 t ti B Gi M l im c cng in trng tng hp trit tiu E M = E 1 + E 2 = 0 M on AB (r 1 = r 2 ) r 1 + r 2 = AB (1) v E 1 = E 2 2 1 2 2 r r = 1 2 q q (2) T (1) v (2) v tr M. b/ Trng hp 2 in tch tri du:( q 1 ,q 2 < 0 ) * 1q > 2q M t ngoi on AB v gn B(r 1 > r 2 ) r 1 - r 2 = AB (1) v E 1 = E 2 2 1 2 2 r r = 1 2 q q (2) 15 16. hongthamvp@gmail.com Su tm v bin son T (1) v (2) v tr M. * 1q < 2q M t ngoi on AB v gn A(r 1 < r 2 ) r 2 - r 1 = AB (1) v E 1 = E 2 2 1 2 2 r r = 1 2 q q (2) T (1) v (2) v tr M. 2/ Tm v tr 2 vect cng in trng do q 1 ,q 2 gy ra ti bng nhau, vung gc nhau: a/ Bng nhau: + q 1 ,q 2 > 0: * Nu 1q > 2q M t ngoi on AB v gn B r 1 - r 2 = AB (1) v E 1 = E 2 2 1 2 2 r r = 1 2 q q (2) * Nu 1q < 2q M t ngoi on AB v gn A(r 1 < r 2 ) r 2 - r 1 = AB (1) v E 1 = E 2 2 1 2 2 r r = 1 2 q q (2) + q 1 ,q 2 < 0 ( q 1 (-); q 2 ( +) M on AB ( nm trong AB) r 1 + r 2 = AB (1) v E 1 = E 2 2 1 2 2 r r = 1 2 q q (2) T (1) v (2) v tr M. b/ Vung gc nhau: r 2 1 + r 2 2 = AB 2 tan = 2 1 E E BI TP VN DNG: Bi 1/ Cho hai in tch im cng du c ln q 1 =4q 2 t ti a,b cch nhau 12cm. im c vect cng in trng do q 1 v q 2 gy ra bng nhau v tr ( s: r 1 = 24cm, r 2 = 12cm) Bi 2/ Cho hai in tch tri du ,c ln in tch bng nhau, t ti A,B cch nhau 12cm .im c vect cng in trng do q 1 v q 2 gy ra bng nhau v tr ( s: r 1 = r 2 = 6cm) Bi 3/ Cho hai in tch q 1 = 9.10 8 C, q 2 = 16.10 8 C t ti A,B cch nhau 5cm . im c vec t cng in trng vung gc vi nhau v E 1 = E 2 ( s: r 1 = 3cm, r 2 = 4cm) Bi 4: Ti ba nh A,B,C ca hnh vung ABCD cnh a = 6cm trong chn khng, t ba in tch im q1=q3= 2.10 -7 C v q2 = -4.10 -7 C. Xc nh in tch q4 t ti D cng in trng tng hp gy bi h in tch ti tm O ca hnhvung bng 0. (q4= -4.10 -7 C) Bi 5: Cho hnh vung ABCD, ti A v C t cc in tch q1=q3=q. Hi phi t B in tch bao nhiu cng in trng D bng khng. (S: q2= 2 2q ) Bi 6: Ti hai nh A,B ca tam gic u ABC cnh a t hai in tch im q1=q2=4.10 -9 C trong khng kh. Hi phi t in tch q3 c gi tr bao nhiu ti C cng in trng gy bi h 3 in tch ti 16 17. hongthamvp@gmail.com Su tm v bin son trng tm G ca tam gic bng0.( q3=4.10 -9 C) Bi 7: Aq1 q2 B 2E u 3E u q3 D C 13E u 1E u : Bn im A, B, C, D trong khng kh to thnh hnh ch nht ABCD cnh AD = a = 3cm, AB = b = 4cm. Cc in tch q1, q2, q3 c t ln lt ti A, B, C. Bit q2=-12,5.10-8 C v cng in trng tng hp ti D bng 0. Tnh q1, q2. Hng dn gii: Vect cng in trng ti D: D 1 3 2 13 2E E E E E E= + + = + u u u u u u V q2 < 0 nn q1, q3 phi l in tch dng. Ta c: 1 2 1 13 2 2 2 q q AD E E cos E cos k k . AD BD BD = = = ( ) 2 3 1 2 232 2 2 AD AD q . q q BD AD AB = = + ( ) 3 8 1 2 2 2 a q .q 2,7.10 a h = = + C Tng t: ( ) 3 8 3 13 2 3 23 2 2 b E E sin E sin q q 6,4.10 C a b = = = = + 1E 2E ---------------------------------------------------------------------------------------------------------------------------- DNG 4:CN BNG CA IN TCH TRONG IN TRNG Bi 1Mt qu cu nh khi lng m=0,1g mang in tch q = 10-8 C c treo bng si dy khng gin v t vo in trng u E c ng sc nm ngang. Khi qu cu cn bng, dy treo hp vi phng thng ng mt gc 0 45 = . Ly g = 10m/s2 . Tnh: a. ln ca cng in trng. b. Tnh lc cng dy . 17 18. hongthamvp@gmail.com Su tm v bin son Hng dn gii: a) Ta co: qE mg.tan 5 tan E 10 V / m mg q = = = Bi 2 in trng gia hai bn ca mt t in phng t nm ngang c cng E = 4900V/m. Xc nh khi lng ca ht bi t trong in trng ny nu n mang in tch q = 4.10-10C v trng thi cn bng. (S: m = 0,2mg) Bi 3: Mt hn bi nh bng kim loi c t trong du. Bi c th tch V=10mm 3 , khi lng m=9.10 -5 kg. Du c khi lng ring D=800kg/m 3 . Tt c c t trong mt in trng u, E hng thng ng t trn xung, E=4,1.10 5 V/m. Tm in tch ca bi n cn bng l lng trong du. Cho g=10m/s2. ( S: q=-2.10-9C) Bi 26: Hai qu cu nh A v B mang nhng in tch ln lt l -2.10 -9 C v 2.10 - 9 C c treo u hai si dy t cch in di bng nhau. Hai im treo M v N cch nhau 2cm; khi cn bng, v tr cc dy treo c dng nh hnh v. Hi a cc dy treo tr v v tr thng ng ngi ta phi dng mt in trng u c hng no v ln bao nhiu? (S: Hng sang phi, E=4,5.10 4 V/m) 18 19. hongthamvp@gmail.com Su tm v bin son ----------------------------------------------------------------------------------------------------------------------------------- --- DNG 5: CNG IN TRNG DO VT TCH IN C KCH THC TO NN ----------------------------------------------------------------------------------------------------------------------------------- --- LUYN TP DNG I: XC NH CNG IN TRNG 19 20. hongthamvp@gmail.com Su tm v bin son Bi 1: in tch im q1=8.10-8 C t ti 0 trong chn khng.Tr li cc cu hi sau: a)xc nh cng in trng ti im cch 0 mt on 30cm. A: 8.103 (V/m); B: 8.102 (V/m); C: 8.104 (V/m); D:800(V/m) b)Nu t q2= -q1 ti M th n chu lc tc dng nh th no? A:Lc ngc chiu CT v c ln 0,64.10-3 N B:Lc cng chiu CT v c ln 0,64.10-3 N Bi 2:mt in tch th t ti im c cng in trng 0,16v/m.lc tc dng ln in tch bng 2.10-4 N.Tnh ln in tch A: 25.10-5 C; B: 125.10-5 C; C:12.10-5 C D:Mt kt qu khc Bi 3:c mt in tch q=5.10-9 C t ti im A trong chn khng.Xc nh cng in trng ti im B cch A mt khong 10cm. A:Hng v A v c ln 4500(v/m); B: Hng ra xa Av c ln 5000(v/m) C:Hng v A v c ln 5000(v/m); D: Hng ra xa A v c ln 4500(v/m) Bi4:Hai in tch q1 =-q2 =10-5 C(q1>0) t 2im A,B(AB=6cm) trong cht in mi c hng s in mi =2. a)Xc nh cng in trng ti im M nm trn ng trung trc ca on AB cch AB mt khong d=4cm A:16.107 V/m; B:2,16..107 V/m; C:2.107 V/m; D: 3.107 V/m. b)xc nh d E t cc i tnh gi tr cc i ca E : A:d=0 v Emax =108 V/m; B:d=10cm v Emax =108 V/m C:d=0 v Emax =2.108 V/m; D: d=10cm v Emax =2.108 V/m Bi 5:cho 2in tch q1=4.10-10 C,q2= -4.10-10 Ct A,B trong khng kh.ChoAB=a=2cm.Xc nh vc t CT E ti cc im sau: a)im H l trung im ca on AB A:72.103 (V/m) B:7200(V/m); C:720(V/m); D:7,2.105 (V/m) b)im M cch A 1cm,ch B3cm. A:32000(V/m); B:320(V/m); C:3200(V/m); D:mt kt qu khc. c)im N hp vi A,B thnh tam gic u A:9000(V/m); B:900(V/m); C:9.104 (V/m); D:mt kt qu khc Bi6:Ti 3 nh A,B,C ca hnh vung ABCD cnh at 3 in tch q ging nhau(q>0).Tnh cng in trng ti cc im sau: a)ti tm 0 ca hnh vung. A:Eo= 2 2 a kq ; B:Eo= 2 2 2 a kq ; C:Eo= 2 2 a qk ; D:E0= a kq2 . b)ti nh D ca hnh vung. A:ED=( 2 + 2 1 ) 2 a kq ; D:ED=2 2 a kq ; C: ED=( 2 +1) 2 a kq ; D:ED=(2+ 2 ) 2 a kq . Bi7:Hai in tch q1=8.10-8 C,q2= -8.10-8 C t ti A,B trong khng kh.AB=4cm.Tm ln vc t ct ti C trn trung trc AB.Cch AB 2cm.suy ra lctc dng ln in tch q=2.10-9 t C A:E=9 2 .105 (V/m) ;F=25,4.10-4 N; B:E=9.105 (V/m) ;F=2.10-4 N. C: E=9000(V/m) ;F=2500N; D:E=900(V/m) ;F=0,002N Bi 8:Ti 2im AvB cch nhau 5cm trong chn khng c 2in tch q1=+16.10-8 c v q2=-9.10-8 c.tnh cng in trng tng hp ti im C nm cch A mt khong 4cm v cch B mt khong 3cm A:12,7.105 (v/m); B;120(v/m); C:1270(v/m) D: mt kt qu khc Bi 9:Ba in tch q ging nhau t ti ba nh ca mt tam gic u cnh a. Xc nh cng in trng ti tm ca tam gic. A:E=0; B:E=1000 V/m; C:E=105 V/m; D: khng xc nh c v cha bit cnh ca tam gic 20 21. hongthamvp@gmail.com Su tm v bin son DNG II: CNG IN TRNG TNG HP BNG KHNG CN BNG IN TCH TRONG IN TRNG Bi 1:Hai in tch im q1=3.10-8 C v q2=-4.10-8 C c t cch nhau ti hai im A,B trong chn khng cch nhau 10cm.hy tm cc im m ti cng in trng bng khng. A: cch A 64,6cm v cch B 74,6cm; B:cch A 64,6cm v cch B 54,6cm; C: cch A 100cm v cch B 110cm; D:cch A 100cm v cch B 90cm Bi 2:Cho hai in tch q1vq2 t A,B trong khng kh.AB=100cm.Tm im C ti cng in trng tng hp bng khng trong cc trng hp sau: a)q1=36.10-6 C; q2=4.10-6 C A: Cch A 75cm v cch B 25cm; B:Cch A25cm v cch B 75cm; C: Cch A 50 cm v cch B 50cm; D: Cch A20cm v cch B 80cm. b)q1=-36.10-6 C;q2=4.10-6 C A: Cch A 50cmv cch B150cm; B:cch B 50cmv cch A150cm; C: cch A 50cm v cch B100cm; D:Cch B50cm v cch A100cm Bi 3:Ti cc nh A v C ca hnh vung ABCD c t cc in tch q1=q3=+q.Hi phi t ti nh B mt in tch q2 bng bao nhiu cng in trng ti D bng khng A: q2= -2 2 .q; B: q2=q; C:q2= -2q; D:q2=2q. Bi 4:Mt qu cu khi lng 1g treo bi si dy mnh trong in trng c cng E=1000V/m c phng ngang th dy treo qu cu lch gc =30o so vi phng thng ng.qu c c in tch q>0(cho g =10m/s2 )Tr li cc cu hi sau: a)Tnh lc cng dy treo qu cu trong in trng A: 3 2 .10-2 N; B: 3 .10-2 N; C: 2 3 .10-2 N; D:2.10-2 N. b)tnh in tch qu cu. A: 3 10 6 C; B: 3 10 5 C ; C: 3 .10-5 C; D: 3 .10-6 C . Bi 5:.Mt qu cu nh khi lng 0,1g c in tch q=10-6 C c treo bngmt si dy mnh trong in trng E=103 V/m c phng ngang cho g=10m/s2 .khi qu cu cn bng,tnh gc lch ca dy treo qu cu so vi phng thng ng. A: 45o ; B:15o ; C: 30o ; D:60o . bi 6:mt ht bi mang in tch dng c khi lng m=10-6 g nm cn bng trong in trng u E c phng nm ngang v c cng E=1000V/m..cho g=10m/s2 ;gc lch ca dy treo so vi phng thng ng l 30o .Tnh in tch ht bi A: 10-9 C; B: 10-12 C; C: 10-11 C; D:10-10 C. Bi 7:Ht bi tch in khi lng m=5mg nm cn bng trong mt in trng u c phng thng ng hng ln c cng E=500 V/m.tnh in tch ht bi(cho g=10m/s2 ) A:10-7 C; B: 10-8 C; C: 10-9 C; D: 2.10-7 C. Bi 8:ti 2 im A v B ch nhau a t cc in tch cng du q1 vq2.Tm c im C trn AB m cng in trng ti C trit tiu.Bit 1 2 q q = n; t CA=x.tnh x(theo a v n) A:x = 1+n a ; B: x = n a ; C:x = n a 1 ; D:x = n a 1+ _______________________________________________________________________________________ ____ - CH 3: IEN THE. HIEU IEN THE. A.L THUYT 21 22. hongthamvp@gmail.com Su tm v bin son 1. Khi mot ien tch dng q dch chuyen trong ien trng eu co cng o E (t M en N) th cong ma lc ien tac dung len q co bieu thc: A = q.E.d Vi: d la khoang cach t iem au iem cuoi (theo phng cua E ). V the d co the dng (d> 0) va cung co the am (d< 0) Cu the nh hnh ve: khi ien tch q di chuyen t M N th d = MH. V cung chieu vi E nen trong trng hp tren d>0. E F Neu A > 0 th lc ien sinh cong dng, A< 0 th lc ien sinh cong am. 2. Cong A ch phu thuoc vao v tr iem au va iem cuoi cua ng i trong ien trng ma khong phu thuoc vao hnh dang ng i. Tnh chat nay cung ung cho ien trng bat k (khong eu). Tuy nhien, cong thc tnh cong se khac. ien trng la mot trng the. 3. The nang cua ien tch q tai mot iem M trong ien trng t le vi o ln cua ien tch q: WM = AM = q.VM. AM la cong cua ien trng trong s dch chuyen cua ien tch q t iem M en vo cc. (moc e tnh the nang.) 4. ien the tai iem M trong ien trng la ai lng ac trng cho kha nang cua ien trng trong viec tao ra the nang cua ien tch q at tai M. 5. Hieu ien the UMN gia hai iem M va N la ai lng ac trng cho kha nang sinh cong cua ien trng trong s di chuyen cua ien tch q t M en N. 6. n v o ien the, hieu ien the la Von (V) II. Hng dan giai bai tap: - Cong ma ta e cap ay la cong cua lc ien hay cong cua ien trng. Cong nay co the co gia tr dng hay am. - Co the ap dung nh ly ong nang cho chuyen ong cua ien tch.Neu ngoai lc ien con co cac lc khac tac dung len ien tch th cong tong cong cua tat ca cac lc tac dung len ien tch bang o tang ong nang cua vat mang ien tch. - Neu vat mang ien chuyen ong eu th cong tong cong bang khong. Cong cua lc ien va cong cua cac lc khac se co o ln bang nhau nhng trai dau. 22 q A q W V MM M == q A VVU MN NMMN == 23. hongthamvp@gmail.com Su tm v bin son - Neu ch co lc ien tac dung len ien tch th cong cua lc ien bang o tang ong nang cua vat mang ien tch. Vi m la khoi lng cua vat mang ien tch q. - Trong cong thc A= q.E.d ch ap dung c cho trng hp ien tch di chuyen trong ien trng eu. III. Bai tap: D NG I : TNH CONG CUA LC IEN. HIEU IEN THE. PP Chung - Cong cua lc ien tac dung len mot ien tch khong phu thuoc vao hnh dang ng i cua ien tch ma ch phu thuoc vao v tr cua iem au va iem cuoi cua ng i trong ien trng. Do o, vi mot ng cong kn th iem au va iem cuoi trung nhau, nen cong cua lc ien trong trng hp nay bang khong. Cong cua lc ien: A = qEd = q.U Cong cua lc ngoai A = A. nh ly ong nang: Bieu thc hieu ien the: q A U MN MN = He thc lien he gia cng o ien trng hieu ien the trong ien trng eu: d U E = 1. Ba iem A, B, C tao thanh mot tam giac vuong tai C. AC = 4 cm, BC = 3 cm va nam trong mot ien trng eu. Vect cng o ien trng E song song vi AC, hng t A C va co o ln E = 5000V/m. Tnh: E a. UAC, UCB, UAB. b. Cong cua ien trng khi mot electron (e) di chuyen t A en B ? s: 200v, 0v, 200v. - 3,2. 10 -17 J. 2. Tam giac ABC vuong tai A c at trong ien trng eu E , = ABC = 60 0 , AB E . Biet BC = 6 cm, UBC= 120V. a. Tm UAC, UBA va cng o ien trng E? E 23 2 . 2 . . 22 MN MNMN vmvm UqA == MNMNMN vvmUqA 22 2 1 . 2 1 . == 24. 1E 2E hongthamvp@gmail.com Su tm v bin son b. at them C ien tch iem q = 9. 10 -10 C. Tm cng o ien trng tong hp tai A. s: UAC = 0V, UBA = 120V, E = 4000 V/m. E = 5000 V/m. 3. Mot ien tch iem q = -4. 10 -8 C di chuyen doc theo chu vi cua mot tam giac MNP, vuong tai P, trong ien trng eu, co cng o 200 v/m. Canh MN = 10 cm, MN E .NP = 8 cm. Moi trng la khong kh. Tnh cong cua lc ien trong cac dch chuyen sau cua q: a. t M N. b. T N P. c. T P M. d. Theo ng kn MNPM. s: AMN= -8. 10 -7 J. ANP= 5,12. 10 -7 J. APM = 2,88. 10 -7 J. AMNPM = 0J. 4. Mot ien trng eu co cng o E = 2500 V/m. Hai iem A , B cach nhau 10 cm khi tnh doc theo ng sc. Tnh cong cua lc ien trng thc hien mot ien tch q khi no di chuyen t A B ngc chieu ng sc. Giai bai toan khi: a. q = - 10 -6 C. b. q = 10 -6 C s: 25. 10 5 J, -25. 10 5 J. 5. Cho 3 ban kim loai phang A, B, C co tch ien va at song song nh hnh. Cho d1 = 5 cm, d2= 8 cm. Coi ien trng gia cac ban la eu va co chieu nh hnh ve. Cng o ien trng tng ng la E1 =4.10 4 V/m , E2 = 5. 10 4 V/m. Tnh ien the cua ban B va ban C neu lay goc ien the la ien the ban A. s: VB = -2000V. VC = 2000V. d1 d2 6. Ba iem A, B, C nam trong ien trng eu sao cho E // CA. Cho AB AC va AB = 6 cm. AC = 8 cm. a. Tnh cng o ien trng E, UAB va UBC. Biet UCD = 100V (D la trung iem cua AC) b. Tnh cong cua lc ien trng khi electron di chuyen t B C, t B D. s: 2500V/m,UAB= 0v, UBC = - 200v. ABC = 3,2. 10 -17 J. ABD= 1,6. 10 - 17 J. 24 25. 1E 2E hongthamvp@gmail.com Su tm v bin son 7. ien tch q = 10 -8 C di chuyen doc theo canh cua mot tam giac eu ABC canh a = 10 cm trong ien trng eu co cng o la 300 V/m. E // BC. Tnh cong cua lc ien trng khi q dch chuyen tren moi canh cua tam giac. s: AAB = - 1,5. 10 -7 J. ABC = 3. 10 -7 J. ACA = -1,5. 10 -7 J. 8. ien tch q = 10 -8 C di chuyen doc theo canh cua mot tam giac eu MBC, moi canh 20 cm at trong ien trng eu E co hng song song vi BC va co cng o la 3000 V/m. Tnh cong thc hien e dch chuyen ien tch q theo cac canh MB, BC va CM cua tam giac. s: AMB = -3J, ABC = 6 J, AMB = -3 J. 9. Gia hai iem B va C cach nhau mot oan 0,2 m co mot ien trng eu vi ng sc hng t B C. Hieu ien the UBC = 12V. Tm: a. Cng o ien trng gia B ca C. b. Cong cua lc ien khi mot ien tch q = 2. 10 -6 C i t B C. s: 60 V/m. 24 J. 10. Cho 3 ban kim loai phang tch ien A, B, C at song song nh hnh. ien trng gia cac ban la ien trng eu va co chieu nh hnh ve. Hai ban A va B cach nhau mot oan d1 = 5 cm, Hai ban B va C cach nhau mot oan d2 = 8 cm. Cng o ien trng tng ng la E1 =400 V/m , d1 d2 E2 = 600 V/m. Chon goc ien the cua ban A. Tnh ien the cua ban B va cua ban C. s: VB = - 20V, VC = 28 V. 11. Mot electron di chuyen c mot oan 1 cm, doc theo mot ng sc ien, di tac dung cua mot lc ien trong mot ien trng eu co cng o 1000 V/m. Hay xac nh cong cua lc ien ? s: 1,6. 10 -18 J. 12. Khi bay t iem M en iem N trong ien trng, electron tang toc, ong nang tang them 250eV.(biet rang 1 eV = 1,6. 10 -19 J). Tm UMN? s: - 250 V. 25 E E 26. O x y b d l 0v E hongthamvp@gmail.com Su tm v bin son CHUYN NG CA IN TCH IM TRONG IN TRNG U A.L THUYT Mt in tch im q dng, khi lng m bay vo in trng u ti im M (in trng u c to bi hai bn kim loi phng rng t song song, i din nhau, hai bn c tch in tri du v bng nhau v ln) vi vn tc ban u 0V u to vi phng ca ng sc in mt gc . Lp phng trnh chuyn ng ca in tch q, Vit phng trnh qu o ca in tch q ri xt cc trng hp ca gc . Cho bit: in trng u c vct cng in trng l E u , M cch bn m mt khong b(m), bn kim loi di l(m), Hai bn cch nhau d(m), gia tc trng trng l g. Li gii: **Chn h trc ta 0xy: Gc 0M. 0x: theo phng ngang(Vung gc vi cc ng sc) 0y: theo phng thng ng t trn xung di (Cng phng, chiu vi ng sc) Gi l gc m vect vn tc ban u ca in Tch hp vi phng thng ng. * Lc tc dng: Trng lc P m.g= u Lc in : F q.E= u Hai lc ny c phng, chiu cng phng chiu vi.ng sc in(Cng phng chiu vi trc 0y) .Phn tch chuyn ng ca q thnh hai chuyn ng thnh phn theo hai trc 0x v 0y. 1. Xt chuyn ng ca q trn phng 0x. Trn phng ny q khng chu bt k mt lc no nn q S chuyn ng thng u trn trc 0x vi vn tc khng i: gia tc ax=0, Vx= V0x =V0. sin (1) =>Phng trnh chuyn ng ca q trn trc 0x: x= Vx.t= V0.. sin .t (2) 2. Xt chuyn ng ca q theo phng 0y: - Theo phng 0y: q chu tc dng ca cc lc khng i(Hp lc cng khng i) q thu c gia tc ay= a = F+P m = q.E g m + u (3) - Vn tc ban u theo phng 0y:V0y= = V0.cos (4) *Vn tc ca q trn trc 0y thi im t l: Vy= V0y+ a.t = V0.cos + ( q.E g m + u ).t (5) 26 27. O x y b d l 0v hongthamvp@gmail.com Su tm v bin son => Phng trnh chuyn ng ca q trn trc 0y: y = V0.cos .t + 1 2 ( q.E g m + u ).t2 (6) TM LI: c im chuyn ng ca q trn cc trc l: Trn trc 0x x x 0 0 a 0 V V .sin x=V .sin .t = = (I) trn trc 0y: y y 0 2 0 q.E a g m q.E V V . os ( g).t m 1 q.E y=V . os .t+ ( g).t 2 m c c = + = + + + (II) ** Phng trnh qu o chuyn ng ca in tch q l( kh t phng trnh ta theo trc 0y bng cch rt t = 0 x V .sin ) y = V0.cos . 0 x V .cos + 1 2 ( q.E g m + u ). 2 0 x ( ) V .sin (7) y = cotg . x + 1 2 . 2 2 0 1 V .sin ( q.E g m + u ). x2 (8) Vy qu o ca q c dng l mt Parabol(Tr nhn gi tr gc 00 , 1800 s nu di) Ch :Bi ton chuyn ng ca e thng b qua trng lc. B.CC DNG BI TP(XT CHO Q>0) DNG 1: VECT VN TC CA IN TCH CNG HNG NG SC a. Gc =0 (Ban u q chuyn ng vo in trng theo hng ca ng sc) Trng hp ny 0V u cng hng vi E u . Da vo (I), (II). Ta c: Trn trc 0x x x 0 0 a 0 V V .sin 0 x=V .sin .t=0 = = = (III) 27 28. hongthamvp@gmail.com Su tm v bin son trn trc 0y: y y 0 0 2 2 0 0 q.E a g m q.E q.E V V . os ( g).t=V ( g).t m m 1 q.E 1 q.E y=V . os .t+ ( g).t V .t+ ( g).t 2 m 2 m c c = + = + + + + + = + (IV) v0 hng cng chiu dng, xt tng hp lc theo 0y, nu n hng cng chiu dng th vt chuyn ng nhanh dn u. 1. Thi gian m q n bn m: khi y= b => b= 2 0 1 q.E V .t+ ( g).t 2 m + -> t. (9) 2. Vn tc khi q p vo bn m l V xc nh theo 2 cch: C1: Thay t (9) vo vo cng thc vn tc ca IV=> V C2: p dng cng thc lin h gia vn tc, gia tc v ng i trong chuyn ng thng nhanh dn u: 2.a.S = V2 - V0 2 tc l 2.a.b = V2 - V0 2 (10) v0 hng cng chiu dng, xt tng hp lc theo 0y, nu n hng ngc chiu dng th vt chuyn ng chm dn u n khi v=0 th chuyn ng nhanh dn u theo hng nguc li. II.BI TP VN DNG Bi 1:Gia 2 bn ca t in t nm ngang cch nhau d=40 cm c mt in trng u E=60V/m. Mt ht bi c khi lng m=3g v in tch q=8.10-5 C bt u chuyn ng t trng thi ngh t bn tch in dng v pha tm tch in m. B qua nh hng ca trng trng. Xc nh vn tc ca ht ti im chnh gia ca t in S:v=0,8m/s Bi 2: Mt electron bay vo trong mt in trng theo hng ngc vi hng ng sc vi vn tc 2000km/s. Vn tc ca electron cui on ng s l bao nhiu nu hiu in th cui on ng l 15V. S:v=3,04.10 6 m/s Bi 3: Mt electron bt u chuyn ng dc theo chiu ng sc in trng ca mt t in phng, hai bn cch nhau mt khong d = 2cm v gia chng c mt hiu in th U = 120V. Electron s c vn tc l bai nhiu sau khi dch chuyn c mt qung ng 1cm. Bi 4: Mt electron bay vo in trng ca mt t in phng theo phng song song cng hng vi cc ng sc in trng vi vn tc ban u l 8.106m/s. Hiu in th t phi c gi tr nh nht l bao nhiu electron khng ti c bn i din S:U>=182V Bi 5: Hi bi c m=10-12 g nm cn bng gia in trng u gia hai bn t.Bit U=125V v d=5cm. a.Tnh in tch ht bi? b.Nu ht bi mt i 5e th mun ht bi cn bng , U=? 28 29. O x y b d l 0v E hongthamvp@gmail.com Su tm v bin son DNG 2: VECT VN TC CA IN TCH NGC HNG NG SC I.L THUYT b. Gc =1800 (Ban u q vo in trng ngc hng ng sc) Trng hp ny 0V u ngc hng vi vc t cng in trngE u . Da vo I, II ta c: Trn trc 0x x x 0 0 a 0 V V .sin 0 x=V .sin .t=0 = = = (V) Trn trc 0y: y y 0 0 2 2 0 0 q.E a g m q.E q.E V V . os ( g).t= - V ( g).t m m 1 q.E 1 q.E y=V . os .t+ ( g).t V .t+ ( g).t 2 m 2 m c c = + = + + + + + = + (VI) Nu tng hp lc in v trng lc trn phng Oy m hng cng Oy th vt chuyn ng theo hai qu trnh. +Qu trnh 1: q chuyn ng thng chm dn u ngc chiu dng trc oy: Gi s: Khi n N th q dng li, qu trnh ny din ra trong thi gian t1 tha mn: 29 30. hongthamvp@gmail.com Su tm v bin son 0 1 q.E - V ( g).t m + + = 0 => t1= 0V q.E g m + . (11) Qung ng MN=S c xc nh: 2.a.S = V2 - V0 2 = - V0 2 (12) (V0 trong trng hp ny ly gi tr m v 0V u ngc hng 0y). * Nu S > d - b th q chuyn ng thng chm dn u ngc chiu dng trc 0y v p vo bn dng gy ra va chm. y a ch xt S < d- b (im N vn nm trong khong khng gian gia hai bn) +Qu trnh 2: Ti N in tch q bt u li chuyn ng thng nhanh dn u theo trc 0y. Vi vn tc ti N bng khng, gia tc a = y q.E a g m = + v bi ton nh trng hp =0. Nu tng hp lc in v trng lc trn phng Oy m ngc hng cng Oy th vt chuyn ng nhanh dn u theo hng ngc Oy. II.BI TP VN DNG 1. Mt e c vn tc ban u vo = 3. 106 m/s chuyn ng dc theo chiu ng sc ca mt in trng c cng in trng E = 1250 V/m. B qua tc dng ca trng trng, e chuyn ng nh th no? s: a = -2,2. 1014 m/s2 , s= 2 cm. 2. Mt e chuyn ng vi vn tc ban u 104 m/s dc theo ng sc ca mt in trng u c mt qung ng 10 cm th dng li. a. Xc nh cng in trng. b. Tnh gia tc ca e. s: 284. 10-5 V/m. 5. 107 m/s2 . 3. Mt e chuyn ng dc theo ng sc ca mt in trng u c cng 364 V/m. e xut pht t im M vi vn tc 3,2. 106 m/s,Hi: a. e i c qung ng di bao nhiu th vn tc ca n bng 0 ? b. Sau bao lu k t lc xut pht e tr v im M ? s: 0,08 m, 0,1 s 4: Mt electron bay t bn m sang bn dng ca mt t in phng. in trng trong khong hai bn t c cng E=6.104 V/m. Khong cch giac hai bn t d =5cm. a. Tnh gia tc ca electron. (1,05.1016 m/s2 ) b. tnh thi gian bay ca electron bit vn tc ban u bng 0.(3ns) c. Tnh vn tc tc thi ca electron khi chm bn dng. (3,2.107 m/s2 ) 5: Gia hai bn kim loi t song song nm ngang tch in tri du c mt hiu in th U1=1000V khong cch gia hai bn l d=1cm. ng gi hai bn c mt git thy ngn nh tch in dng nm l lng. t nhin hiu in th gim xung ch cn U2 = 995V. Hi sau bao lu git thy ngn ri xung bn dng? DNG 3: VECT VN TC CA IN TCH VUNG GC NG SC I.L THUYT c. Gc =900 (Ban u q bay vo theo hng vung gc vi ng sc in) Da vo I, II ta c: Trn trc 0x (VI) 30 31. hongthamvp@gmail.com Su tm v bin son Trn trc 0y: y y 0 2 2 0 q.E a g m q.E q.E V V . os ( g).t= ( g).t m m 1 q.E 1 q.E y=V . os .t+ ( g).t ( g).t 2 m 2 m c c = + = + + + + = + (VII) T trn ta khng nh q chuyn ng nh chuyn ngca vt b nm ngang. Thi gian q n c bn m l t1 tha mn: y = b b = 2 1 1 q.E ( g).t 2 m + => t1 (13) kim tra xem q c p vo bn m khng ta phi xt: x = 0 1V .t l (14) II.BI TP VN DNG Bi1. Mt e c bn vi vn tc u 2. 10-6 m/s vo mt in trng u theo phng vung gc vi ng sc in. Cng in trng l 100 V/m. Tnh vn tc ca e khi n chuyn ng c 10-7 s trong in trng. in tch ca e l 1,6. 10-19 C, khi lng ca e l 9,1. 10-31 kg. s: F = 1,6. 10-17 N. a = 1,76. 1013 m/s2 vy = 1, 76. 106 m/s, v = 2,66. 106 m/s. Bi 2. Mt e c bn vi vn tc u 4. 107 m/s vo mt in trng u theo phng vung gc vi cc ng sc in. Cng in trng l 103 V/m. Tnh: a. Gia tc ca e. b. Vn tc ca e khi n chuyn ng c 2. 10-7 s trong in trng. s: 3,52. 1014 m/s2 . 8,1. 107 m/s. Bi 3. . Cho 2 bn kim loi phng c di l=5 cm t nm ngang song song vi nhau,cch nhau d=2 cm. Hiu in th gia 2 bn l 910V. Mt e bay theo phng ngang vo gia 2 bn vi vn tc ban u v0=5.107 m/s. Bit e ra khi c in trng. B qua tc dng ca trng trng 1) Vit ptrnh qu o ca e trong in trng(y=0,64x2 ) 2) Tnh thi gian e i trong in trng? Vn tc ca n ti im bt u ra khi in trng?(10-7 s, 5,94m/s) 3) Tnh lch ca e khi phng ban u khi ra khi in trng? ( S: 0,4 cm) Bi 4: Mt electron bay trong in trng gia hai bn ca mt t in tch in v t cch nhau 2cm vi vn tc 3.107 m/s theo phng song song vi cc bn ca t in. Hiu in th gia hai bn phi l bao nhiu electron lch i 2,5mm khi i c on ng 5cm trong in trng. Bi 5.Sau khi c tng tc bi U=200V, mt in t bay vo chnh gia hai bn t theo phng song song hai bn.Hai bn c chiu di l=10cm, khong cch gia hai bn d=1cm.Tm U gia hai bn in t khng ra khi uc t? S: U>=2V Bi 6.Mt e c ng nng 11,375eV bt u vo in trng u nm gia hai bn theo phng vung gc vi ng sc v cch u hai bn. a.Tnh vn tc v0 lc bt u vo in trng? b,Thi gian i ht l=5cm ca bn. c. dch theo phng thng ng khi e ra khi in trng, bit U=50V, d=10cm. d.ng nng v vn tc e ti cui bn Bi 7.in t mang nng lng 1500eV bay vo t phng theo hng song song hai bn.Hai bn di l=5cm, cch nhau d=1cm.Tnh U gia hai bn in t bay ra khi t theo phng hp cc bn gc 110 . 31 32. hongthamvp@gmail.com Su tm v bin son S:U=120V DNG 4: VECT VN TC CA IN TCH XIN GC NG SC d. Trng hp gc 900 < < 1800 th in tch q chuyn ng nh mt vt b nm xin ln. Ta ca nh Parabol l: (Da theo cng thc y = cotg . x + 1 2 . 2 2 0 1 V .sin . x2 ) x= 2 2 0 0 2 2 0 cotg 2V cos .sin V sin2 1 1 . 2 V .sin = = (15) y= - V0 2 . cos2 + 1 2 .V0 2. .4.cos2 = V0 2 . cos2 . (16) Xt xem q c p dng hay khng: Xem ta nh:y>b-d th c v ngc li th khng Xt xem q c p vo bn m hay khng: Thi gian q c ta y = b l t tha mn phng trnh (13) Kim tra xem khi x< l hay cha . e. Trng hp 00 < < 900 th q chuyn ng nh vt b nm xin xung. Ta nh ca Parabol l x=0, y=0. q p vo bn m thi im t1 tha mn y = b. (Nu x(t1) > l th q bay ra ngoi m khng p vo bn m cht no) Thng l x(t1) < l nn q p vo bn m ti im K . K cch mp tri bn m khong x(t1). II.BI TP VN DNG Bi 1: Hai bn kim loi ni vi ngun in khng i c hiu in th U = 228 V. Ht electron c vn tc ban u v 0 = 4.107 m/s, bay vo khong khng gian gia hai bn qua l nh O bn dng, theo phng hp vi bn dng gc 0 60 = . a, Tm qu o ca electron sau . b, Tnh khong cch h gn bn m nht m electron t ti, b qua tc dng ca trng lc . Bi 2:Hai bn kim loi tch in tri du t cch nhau d=3cm, chiu di mi bn l=5cm. Mt in t lt vo gia hai bn hp bn dng gc 300 . Xc nh U sao cho khi chui ra khi bn in t chuyn ng theo phng song song vi hai bn? S: U=47,9V 32 33. hongthamvp@gmail.com Su tm v bin son _______________________________________________________________________________________ _____ CH 4: BI TP V T IN DNG I:TNH TON CC I LNG PP Chung: Van dung cong thc: ien dung cua tu ien: U Q C = (1) Nang lng cua tu ien: 2 2 . 2 1 . 2 1 2 1 UCU C Q W === Q ien dung cua tu ien phang: d S d S C o ..4.10.9 ... 9 == (2) Trong o S la dien tch cua mot ban (la phan oi dien vi ban kia) oi vi tu ien bien thien th phan oi dien cua hai ban se thay oi. Cong thc (2) ch ap dung cho trng hp chat ien moi lap ay khoang khong gian gia hai ban. Neu lp ien moi ch chiem mot phan khoang khong gian gia hai ban th can phai phan tch, lap luan mi tnh c ien dung C cua tu ien. - Lu y cac ieu kien sau: + Noi tu ien vao nguon: U = const. + Ngat tu ien khoi nguon: Q = const. 1. Tu ien phang gom hai ban tu co dien tch 0,05 m 2 at cach nhau 0,5 mm, ien dung cua tu la 3 nF. Tnh hang so ien moi cua lp ien moi gia hai ban tu. s: 3,4. 2. Mot tu ien khong kh neu c tch ien lng 5,2. 10 -9 C th ien trng gia hai ban tu la 20000 V/m. Tnh dien tch moi ban tu. s: 0,03 m 2 . 3. Mot tu ien phang ien dung 12 pF, ien moi la khong kh. Khoang cach gia hai ban tu 0,5 cm. Tch ien cho tu ien di hieu ien the 20 V. Tnh: a. ien tch cua tu ien. b. Cng o ien trng trong tu. s: 24. 10 -11 C, 4000 V/m. 4. Mot tu ien phang khong kh, ien dung 40 pF, tch ien cho tu ien hieu ien the 120V. a. Tnh ien tch cua tu. 33 34. hongthamvp@gmail.com Su tm v bin son b. Sau o thao bo nguon ien roi tang khoang cach gia hai ban tu len gap oi. Tnh hieu ien the mi gia hai ban tu. Biet rang ien dung cua tu ien phang t le nghch vi khoang cach gia hai ban cua no. s: 48. 10 -10 C, 240 V. 5. Tu ien phang khong kh co ien dung C = 500 pF c tch ien en hieu ien the 300 V. a. Tnh ien tch Q cua tu ien. b. Ngat tu ien khoi nguon roi nhung tu ien vao chat ien moi long co = 2. Tnh ien dung C1 , ien tch Q1 va hieu ien the U1 cua tu ien luc o. c. Van noi tu ien vi nguon nhng nhung tu ien vao chat ien moi long co = 2. Tnh C2 , Q2 , U2 cua tu ien. s: a/ 150 nC ; b/ C1 = 1000 pF, Q1 = 150 nC, U1 = 150 V. c/ C2 = 1000 pF, Q2 = 300 nC, U2 = 300 V. 6. Tu ien phang khong kh ien dung 2 pF c tch ien hieu ien the 600V. a. Tnh ien tch Q cua tu. b. Ngat tu khoi nguon, a hai au tu ra xa e khoang cach tang gap oi. Tnh C1, Q1, U1 cua tu. c. Van noi tu vi nguon, a hai ban tu ra xa e khoang cach tang gap oi. Tnh C2, Q2, U2 cua tu. s: a/ 1,2. 10 -9 C. b/ C1 = 1pF, Q1 = 1,2. 10 -9 C, U1 = 1200V. c/ C2 = 1 pF, Q2 = 0,6. 10 -9 C, U 2 = 600 V. 7 Tu ien phang co cac ban tu hnh tron ban knh 10 cm. Khoang cach va hieu ien the gia hai ban la 1cm, 108 V. Gia hai ban la khong kh. Tm ien tch cua tu ien ? s: 3. 10 -9 C. 8. Tu ien phang gom hai ban tu hnh vuong cach a = 20 cm at cach nhau 1 cm. Chat ien moi gia hai ban la thuy tinh co = 6. Hieu ien the gia hai ban U = 50 V. 34 35. hongthamvp@gmail.com Su tm v bin son a. Tnh ien dung cua tu ien. b. Tnh ien tch cua tu ien. c. Tnh nang lng cua tu ien, tu ien co dung e lam nguon ien c khong ? s: 212,4 pF ; 10,6 nC ; 266 nJ. 9.T in cu to bi qu cu bn knh R1v v cu bn knh R2(R1< R2).Tnh in dung ca qu cu ny? S: 1 2 1 2( ) R R k R R+ DNG II:GHP T CHA TCH IN A.L THUYT PP Chung: - Van dung cac cong thc tm ien dung (C), ien tch (Q), hieu ien the (U) cua tu ien trong cac cach mac song song, noi tiep. - Neu trong bai toan co nhieu tu c mac hon hp, ta can tm ra c cach mac tu ien cua mach o roi mi tnh toan. - Khi tu ien b anh thung, no tr thanh vat dan. - Sau khi ngat tu ien khoi nguon va van gi tu ien o co lap th ien tch Q cua tu o van khong thay oi. oi vi bai toan ghep tu ien can lu y hai trng hp: + Neu ban au cac tu cha tch ien, khi ghep noi tiep th cac tu ien co cung ien tch va khi ghep song song cac tu ien co cung mot hieu ien the. + Neu ban au tu ien (mot hoac mot so tu ien trong bo) a c tch ien can ap dung nh luat bao toan ien tch (Tong ai so cac ien tch cua hai ban noi vi nhau bang day dan c bao toan, ngha la tong ien tch cua hai ban o trc khi noi vi nhau bang tong ien tch cua chung sau khi noi). . Nghin cu v s thay i in dung ca t in phng + Khi a mt tm in mi vo bn trong t in phng th chnh tm l mt t phng v trong phn cp phn in tch i din cn li to thnh mt t iin phng. Ton b s to thnh mt mch t m ta d dng tnh in dung. in dung ca mch chnh l in dung ca t khi thay i in mi. + Trong t in xoay c s thay i in dung l do s thay i in tch i din ca cc tm. Nu l c n tm th s c (n-1) t phng mc song song. B.BI TP VN DNG 1. Mot tu ien phang ien dung C = 0,12 F co lp ien moi day 0,2 mm co hang so ien moi = 5. Tu c at di mot hieu ien the U = 100 V. a. Tnh dien tch cac ban cua tu ien, ien tch va nang lng cua tu. 35 36. hongthamvp@gmail.com Su tm v bin son b. Sau khi c tch ien, ngat tu khoi nguon roi mac vao hai ban cua tu ien C1 = 0,15 F cha c tch ien. Tnh ien tch cua bo tu ien, hieu ien the va nang lng cua bo tu. s: a/ 0,54 m 2 , 12 C, 0,6 mJ. b/ 12 C, 44,4 V, 0,27 mJ. 2. Mot tu ien 6 F c tch ien di mot hieu ien the 12V. a. Tnh ien tch cua moi ban tu. b. Hoi tu ien tch luy mot nang lng cc ai la bao nhieu ? c. Tnh cong trung bnh ma nguon ien thc hien e a 1 e t ban mang ien tch dng ban mang ien tch am ? s: a/ 7,2. 10 -5 C. b/ 4,32. 10 -4 J. c/ 9,6. 10 -19 J. 3. Tnh ien dung tng ng, ien tch, hieu ien the trong moi tu ien cac trng hp sau (hnh ve) C2 C3 C2 C1 C2 C3 C1 C2 C3 C1 C1 C3 (Hnh 1) (Hnh 2) (Hnh 3) (Hnh 4) Hnh 1: C1 = 2 F, C2 = 4 F, C3 = 6 F. UAB = 100 V. Hnh 2: C1 = 1 F, C2 = 1,5 F, C3 = 3 F. UAB = 120 V. Hnh 3: C1 = 0,25 F, C2 = 1 F, C3 = 3 F. UAB = 12 V. Hnh 4: C1 = C2 = 2 F, C3 = 1 F, UAB = 10 V. 4. Co 3 tu ien C1 = 10 F, C2 = 5 F, C3 = 4 F c mac vao nguon ien co C1 C3 hieu ien the U = 38 V. a. Tnh ien dung C cua bo tu ien, ien tch va hieu ien the tren cac C2 tu ien. b. Tu C3 b anh thung. Tm ien tch va hieu ien the tren tu C1. s: a/ Cb 3,16 F. Q1 = 8. 10 -5 C, Q2 = 4. 10 -5 C, Q3 = 1,2. 10 -4 C, 36 37. hongthamvp@gmail.com Su tm v bin son U1 = U2 = 8 V, U3 = 30 V. b/ Q1 = 3,8. 10 -4 C, U1 = 38 V. 5. Cho bo tu mac nh hnh ve: C1 = 1 F, C2 = 3 F, C3 = 6 F, C4 = 4 F. UAB = 20 V. C1 C2 Tnh ien dung bo tu, ien tch va hieu ien the moi tu khi. a. K h. C3 C4 b. K ong. 6. Trong hnh ben C1 = 3 F, C2 = 6 F, C3 = C4 = 4 F, C5 = 8 F. C1 C2 U = 900 V. Tnh hieu ien the gia A va B ? C3 C4 s: UAB = - 100V. C5 7. Cho mach ien nh hnh ve: C1 = C2 = C3 = C4 =C5 = 1 F, U = 15 V. C1 C2 Tnh ien dung cua bo tu, ien tch va hieu ien the cua moi tu khi: C5 a. K h. b. K ong. C3 C4 8. Cho bo tu ien nh hnh ve. C2 C2 C2 = 2 C1, UAB = 16 V. Tnh UMB. C1 C1 C1 s: 4 V. 9. Cho bo 4 tu ien giong nhau ghep theo 2 cach nh hnh ve. a. Cach nao co ien dung ln hn. b. Neu ien dung tu khac nhau th chung phai co lien he the nao e CA = CB (ien dung cua hai cach ghep bang nhau) Hnh A. 37 38. hongthamvp@gmail.com Su tm v bin son Hnh B. s: a/ CA = 3 4 CB. b/ 21 21 4 . CC CC C + = Bi 10:Tnh in dung ca b t in, in tch v hiu in th ca mi t trong cc trng hp sau y: a) C1=2 F ; C2=4 F C3=6 F ; U= 100V b) C1=1 F ; C2=1,5 F C3=3 F ; U= 120V c) C1=0,25 F ; C2=1 F C3=3 F ; U= 12V /S :C=12 F ;U1=U2=U3= 100V Q1=2.10-4C; Q2= 4.10-4C Q3= 6.10-4C /S :C=0,5 F ;U1=60V;U2=40V;U3= 20V Q1= Q2= Q3= 6.10-5C /S: C=1 F ;U1=12V;U2=9V U3= 3V Q1=3.10-6 C; Q2=Q3= 910-6 C Bi11:: Hai t in khng kh phng c in dung l C1= 0,2 F v C2= 0,4 F mc song song. B c tch in n hiu in th U=450V ri ngt khi ngun. Sau lp y khong gia hai bn t in C2 bng in mi c hng s in mi l 2. Tnh in th ca b t v in tch ca mi t /S: 270V; 5,4.10-5 C v 2,16. 10-5 C Bi12: Hai t in phng c C1= 2C2,mc ni tip vo ngun U khng i. Cng in trng trong C1 thay i bao nhiu ln nu nhng C2 vo cht in mi c 2= . /S: Tng 1,5 ln Bi 13: Ba tm kim loi phng ging nhau t song song vi nhau nh hnh v: Din tch ca mi bn l S= 100cm2, Khong cch gia hai bn lin tip l d= 0,5cm Ni A v B vi ngun U= 100V a) Tnh in dung ca b t v in tch ca mi bn b) Ngt A v B ra khi ngun in. Dch chuyn bn B theo phng vung gc vi cc bn t in mt on l x. Tnh hiu in th gia A v B theo x. p dng khi x= d/2 /s: a) 3,54.10-11 F; 1,77.10-9 C v 3,54.10-9 C b) 2 22 ' . d xd UU = ; 75V Bi 14: Bn tm kim loi phng ging nhau nh hnh v. Khong cch BD= 2AB=2DE. B v D c ni vi ngun in U=12V, sau ngt ngun i. Tm hiu in th gia B v D nu sau : a) Ni A vi B b) Khng ni A vi B nhng lp y khong gia B v D bng in mi 3= /S a) 8V b) 6V Bi 15: T in phng khng kh C=2pF. Nhng chm mt na vo trong in mi lng 3= . Tm in dung ca t in nu khi nhng, cc bn t : a) Thng ng b) Nm ngang 38 C1 C2 C3 C1 C2 C3 C1 C2 C3 A B A B D C 39. hongthamvp@gmail.com Su tm v bin son /S a) 4pF b)3pF Cho b t in mc nh hnh v bn: Chng minh rng nu c: 4 3 2 1 C C C C = Hoc 4 2 3 1 C C C C = Th khi K ng hay K m, in dung ca b t u khng thay i Bi 16 Mt t in phng c in dung C0. Tm in dung ca t in khi a vo bn trong t mt tm in mi c hng s in mi , c din tch i din bng mt na din tch mt tm, c chiu dy bng mt phn ba khong cch hai tm t, c b rng bng b rng tm t, trong hai trng hp sau: DNG III:GHP T CHA IN TCH Bi 1: em tch in cho t in C1 = 3 F n hiu in th U1 = 300V, cho t in C2 = 2 F n hiu in th U2 = 220V ri: a) Ni cc tm tch in cng du vi nhau b) Ni cc tm tch in khc du vi nhau c) Mc ni tip hai t in (hai bn m c ni vi nhau) ri mc vo hiu in th U = 400V. Tm in tch v hiu in th ca mi t trong trong trng hp trn. Bi 2: em tch in cho t in C1 = 1 F n hiu in th U1 = 20V, cho t in C2 = 2 F n hiu in th U2 = 9V .Sau ni hai bn m hai t vi nhau, 2 bn dng ni vi hai bn ca t C3=3 F cha tch in. a.Tnh in tch v hiu in th mi bn sau khi ni? b.Xc nh chiu v s e di chuyn qua dy ni hai bn m hai t C1 v C2? . . DNG IV:HIU IN TH GII HN A.L THUYT Trng hp 1 t:Ugh=Egh.d Trng hp nhiu t: Ub=Min(Uigh) B.BI TP 39 A B M N C4 C2 C1 C3 K 40. hongthamvp@gmail.com Su tm v bin son Bi 1: Hai bn ca mt t in phng c dng hnh trn bn knh R = 30cm, khong cch gia hai bn l d = 5mm, gia hai bn l khng kh. a. Tnh in dung ca t. b. Bit rng khng kh ch cch in khi cng in trng ti a l 3.105 V/m. Hi: - Hiu in th gii hn ca t in. - C th tch cho t in mt in tch ln nht l bao nhiu t khng b nh thng? Bi 2: hai t in c in dung ln lt C1 = 5.10-10 F v C2 = 15.10-10 F, c mc ni tip vi nhau. Khong cch gia hai bn ca mi t in l d = 2mm. in trng gii hn ca mi t Egh = 1800V. Tnh hiu in th gii hn ca b t. Ugh=4,8V Bi 3 Ba t in c in dung C1=0,002 F; C2=0,004 F; C3=0,006 F c mc ni tip thnh b. Hiu in th nh thng ca mi t in l 4000 V.Hi b t in trn c th chu c hiu in th U=11000 V khng? Khi hiu in th t trn mi t l bao nhiu? S: Khng. B s b nh thng; U1=6000 V; U2=3000 V; U3=2000 V Bi 4 Mt b t gm 5 t in ging ht nhau ni tip mi t c C=10 F c ni vo ht 100 V 1) Hi nng lng ca b thay i ra sao nu 1 t b nh thng 2) Khi t trn b nh thng th nng lng ca b t b tiu hao do phng in. Tm nng lng tiu hao . Bi 5: Hai t c C1=5F, C2=10F; Ugh1=500V, Ugh2=1000V;.Ghp hai t in thnh b. Tm hiu in th gii hn ca b t in nu hai t: a.Ghp song song b.Ghp ni tip DNG V:T C CHA NGUN,T XOAY 1. Cho mch nh hnh v. Bit C1=2F, C2=10F, C3=5F; U1=18V, U2=10V. Tnh in tch v HT trn mi t? C1 M C2 C3 + U1- N - U2 + 2.Cho mach nh hnh v. Bit U1=12V, U2=24V; C1=1F, C2=3F. Lc u kho K m. a/ Tnh in tch v HT trn mi t? b/ Kho K ng li. Tnh in lng qua kho K C1 M C2 K + U1- N + U2 3. Cho mch nh hnh v: Bit C1=1F, C2=3F, C3=4F, C4=2; U=24V. a/ Tnh in tch cc t khi K m? b/Tm in lng qua kho K khi K ng. C1 M C2 A K B 40 41. hongthamvp@gmail.com Su tm v bin son Bi4Cho mt s t in ging nhau c in dung l C0= 3 F . Nu cch mc dng t nht cc t in trn mc thnh b t c in dung l C= 5 F . V s cch mc ny? Bi 5: Cho b t nh hnh v .Tnh in dung ca b t hiu in th gia hai bn t in, v in tch ca cc t. Cho bit: C1=C3=C5=1 F ; C2= 4 F ; v C4= 1,2 F . U= 30V Bi 6: Cho b t in nh hnh v sau y: C2= 2C1; UAB= 16V. Tnh UMB C3 C4 Bi 7: Cho mch t nh hnh, bit: C1 = 6 F, C2 =4 F, C3 = 8 F, C4 = 5 F, C5 = 2 F. Hy tnh in dung ca b + U . T XOAY: Bi 1:T xoay gm n tm hnh bn nguyt ng knh D=12cm, khong cch gia hai tm lin tip d=0,5mm. Phn i din gia hai bn c nh v bn di chuyn c dng hnh qut vi gc tm l 00 0, t gia hai in tch cch 4q khong r/3 B. Q < 0, t gia hai in tch cch 4q khong 2r/3 C.Q tri du vi q t gia 2 in tch cch q khong r/3 D.Q ty t gia 2 in tch cch q khong r/3 Cu 3: Ti bn nh ca mt hnh vung t 4 in tch im ging nhau q = + 1C v ti tm hnh vung t in tch q0, h nm in tch cn bng. Tm du v ln in tch im q0? A. q0 = + 0,96 C B. q0 = - 0,76 C C. q0 = + 0,36 C D. q0 = - 0,96 C Cu 4: Mt qu cu khi lng 10g mang in tch q1 = + 0,1C treo vo mt si ch cch in, ngi ta a qu cu 2 mang in tch q2 li gn th qu cu th nht lch khi v tr ban u mt gc 300 , khi hai qu cu trn cng mt mt phng nm ngang cch nhau 3cm. Tm du, ln in tch q2 v sc cng ca si dy: A. q2 = + 0,087 C B. q2 = - 0,087 C C. q2 = + 0,17 C D. q2 = - 0,17 C Cu 5: Ngi ta treo hai qu cu nh khi lng bng nhau m = 0,01g bng hai si dy c di nh nhau l = 50cm( khi lng khng ng k). Cho chng nhim in bng nhau chng y nhau cch nhau 6cm. Tnh in tch mi qu cu: A. q = 12,7pC B. q = 19,5pC C. q = 15,5nC D.q = 15,5.10-10 C Cu 6: Treo hai qu cu nh khi lng bng nhau m bng nhng si dy cng di l( khi lng khng ng k). Cho chng nhim in bng nhau chng y nhau cch nhau khong r = 6cm. Nhng c h thng vo trong ru c = 27, b qua lc y Acsimet, tnh khong cch gia chng khi tng tc trong du: A. 2cm B. 4cm C. 6cm D. 1,6cm Cu 7: Ngi ta treo hai qu cu nh khi lng bng nhau m = 0,1g bng hai si dy c di nh nhau l ( khi lng khng ng k). Cho chng nhim in bng nhau chng y nhau v cn bng khi mi dy treo hp vi phng thng ng mt gc 150 . Tnh lc tng tc in gia hai qu cu: A. 26.10-5 N B. 52.10-5 N C. 2,6.10-5 N D. 5,2.10-5 N Cu 8: Ngi ta treo hai qu cu nh khi lng bng nhau m = 0,1g bng hai si dy c di nh nhau l = 10cm( khi lng khng ng k). Truyn mt in tch Q cho hai qu cu th chng y nhau cn bng khi mi dy treo hp vi phng thng ng mt gc 150 , ly g = 10m/s2 . Tnh in tch Q: A. 7,7nC B. 17,7nC C. 21nC D. 27nC Cu 9: Ba in tch bng nhau q dng t ti 3 nh ca tam gic u ABC cnh a. Hi phi t mt in tch q0 nh th no v u lc in tc dng ln cc in tch cn bng nhau: 49 50. hongthamvp@gmail.com Su tm v bin son A. q0 = +q/ 3 , gia AB B. q0 = - q/ 2 , trng tm ca tam gic C. q0 = - q/ 3 , trng tm ca tam gic D. q0 = +q/ 3 , nh A ca tam gic Cu 10: Hai qu cu nh bng kim loi ging ht nhau tch in dng treo trn hai si dy mnh cng chiu di vo cng mt im. Khi h cn bng th gc hp bi hai dy treo l 2. Sau cho chng tip xc vi nhau ri bung ra, chng cn bng th gc lch by gi l 2 '. So snh v ': A. > ' B. < ' C. = ' D. c th ln hoc nh hn ' Cu 1 2 3 4 5 6 7 8 9 10 p n D C D B D A A B C B in tch, Fculng - Dng 3: in tch cn bng chu td lc Culng - 2: Cu 1: Hai qu cu nh kim loi ging ht nhau mang in tch q1 v q2 t trong chn khng cch nhau 20cm ht nhau mt lc 5.10- 7 N. t vo gia hai qu cu mt tm thy tinh dy d = 5cm c hng s in mi = 4 th lc lc ny tng tc gia hai qu cu l bao nhiu? A. 1,2.10-7 N B. 2,2.10-7 N C. 3,2.10-7 N D.4 ,2.10-7 N Cu 2: Hai qu cu ging nhau khi lng ring l D tch in nh nhau treo u ca hai si dy di nh nhau t trong du khi lng ring D0, hng s in mi = 4 th gc lch gia hai dy treo l . Khi t ra ngoi khng kh thy gc lch gia chng vn bng . Tnh t s D/ D0 A. 1/2 B. 2/3 C. 5/2 D. 4/3 Cu 3: Bn in tch im q1, q2, q3, q4 t trong khng kh ln lt ti cc nh ABCD ca hnh vung thy hp lc tnh in tc dng ln q4 ti D bng khng. Gia 3 in tch kia quan h vi nhau: A. q1 = q3; q2 = q1 2 B. q1 = - q3; q2 = ( 1+ 2 )q1 C. q1 = q3; q2 = - 2 2 q1 D. q1 = - q3; q2 = ( 1- 2 )q1 Cu 4: Hai in tch im trong khng kh q1 v q2 = - 4q1 ti A v B, t q3 ti C th hp cc lc in tc dng ln q3 bng khng. Hi im C c v tr u: A. trn trung trc ca AB B. Bn trong on AB C. Ngoi on AB. D. khng xc nh c v cha bit gi tr ca q3 Cu 5: Hai in tch im trong khng kh q1 v q2 = - 4q1 ti A v B vi AB = l, t q3 ti C th hp cc lc in tc dng ln q3 bng khng. Khong cch t A v B ti C ln lt c gi tr: A. l/3; 4l/3 B. l/2; 3l/2 C. l; 2l D. khng xc nh c v cha bit gi tr ca q3 Cu 6: Hai qu cu kim loi nh ging nhau khi lng m, tch in cng loi bng nhau c treo bi hai si dy nh di l cch in nh nhau vo cng mt im. Chng y nhau khi cn bng hai qu cu cch nhau mt on r 0; q2 < 0; |q1| = |q2| C. q1 < 0; q2 < 0; q1 = q2 D. q1 < 0; q2 >0; |q1| = |q2| Cu hi 3: Hai in tch t trong khng kh ti M v N. Ti I nm trn ng trung trc ca MN cch MN mt on IH c vct cng in trng tng hp IE nm theo ng trung trc IH v hng li gn MN th hai in tch c c im: A. q1 > 0; q2 > 0; q1 = q2 B. q1 > 0; q2 < 0; |q1| = |q2| C. q1 < 0; q2 < 0; q1 = q2 D. q1 < 0; q2 >0; |q1| = |q2| Cu hi 4: Hai in tch t trong khng kh ti M v N. Ti I nm trn ng trung trc ca MN cch MN mt on IH c vct cng in trng tng hp IE song song vi MN th hai in tch c c im: A. q1 > 0; q2 > 0; q1 = q2 B. q1 > 0; q2 < 0; |q1| = |q2| C. q1 < 0; q2 >0; |q1| = |q2| D. B hoc C Cu hi 5: Hai in tch q1 = +q v q2 = - q t ti A v B trong khng kh, bit AB = 2a. ln cng in trng ti M trn ng trung trc ca AB cch AB mt on h l: A. 22 2 ha kq + B. ( )222 2 2 ha kqa + C. ( )2 3 22 2 ha kqa + D. 22 2 2 ha kqa + Cu hi 6: Hai in tch q1 = +q v q2 = - q t ti A v B trong khng kh, bit AB = 2a. ti M trn ng trung trc ca AB cch AB mt on h EM c gi tr cc i. Gi tr cc i l: A. 2 2a kq B. 2 a kq C. 2 2 a kq D. 2 4 a kq Cu hi 7: Ba in tch q1, q2, q3 t trong khng kh ln lt ti cc nh A, B, C ca hnh vung ABCD. Bit vct cng in trng tng hp ti D c gi l cnh CD. Quan h gia 3 in tch trn l: A. q1 = q2 = q3 B. q1 = - q2 = q3 C. q2 = - 2 2 q1 D. q3 = - 2 2 q2 Cu hi 8: Hai in tch im q1 = 2.10-2 (C) v q2 = - 2.10-2 (C) t ti hai im A v B cch nhau mt on a = 30 (cm) trong khng kh. Cng in trng ti im M cch u A v B mt khong bng a c ln l: A. EM = 0,2 (V/m). B. EM = 1732 (V/m). C. EM = 3464 (V/m). D. EM = 2000 (V/m). Cu hi 9: Hai in tch q1 = 5.10-16 (C), q2 = - 5.10-16 (C), t ti hai nh B v C ca mt tam gic u ABC cnh bng 8 (cm) trong khng kh. Cng in trng ti nh A ca tam gic ABC c ln l: A. E = 1,2178.10-3 (V/m). B. E = 0,6089.10-3 (V/m). C. E = 0,3515.10-3 (V/m). D. E = 0,7031.10-3 (V/m). 54 55. hongthamvp@gmail.com Su tm v bin son Cu hi 10: Hai in tch im q1 = 0,5 (nC) v q2 = - 0,5 (nC) t ti hai im A, B cch nhau 6 (cm) trong khng kh. Cng in trng ti trung im ca AB c ln l: A. E = 0 (V/m). B. E = 5000 (V/m). C. E = 10000 (V/m). D. E = 20000 (V/m). Cu 1 2 3 4 5 6 7 8 9 10 p n D A C D C A C B D C in trng - Dng 3: q cn bng trong in trng, E trit tiu - 1 Cu hi 1: Hai in tch im q1 v q2 t ti hai im c nh A v B. Ti im M trn ng thng ni AB v gn A hn B ngi ta thy in trng ti c cng bng khng. Kt lun g v q1 , q2: A. q1 v q2 cng du, |q1| > |q2| B. q1 v q2 tri du, |q1| > |q2| C. q1 v q2 cng du, |q1| < |q2| D. q1 v q2 tri du, |q1| < |q2| Cu hi 2: Hai in tch im q1 = - 9C, q2 = 4 C t ln lt ti A, B cch nhau 20cm. Tm v tr im M ti in trng bng khng: A. M nm trn on thng AB, gia AB, cch B 8cm B. M nm trn ng thng AB, ngoi gn B cch B 40cm C. M nm trn ng thng AB, ngoi gn A cch A 40cm D. M l trung im ca AB Cu hi 3: Hai in tch im q1 = - 4 C, q2 = 1 C t ln lt ti A v B cch nhau 8cm. Xc nh v tr im M ti cng in trng bng khng: A. M nm trn AB, cch A 10cm, cch B 18cm B. M nm trn AB, cch A 8cm, cch B 16cm C. M nm trn AB, cch A 18cm, cch B 10cm D. M nm trn AB, cch A 16cm, cch B 8cm Cu hi 4: Hai tm kim loi phng nm ngang nhim in tri du t trong du, in trng gia hai bn l in trng u hng t trn xung di v c cng 20 000V/m. Mt qu cu bng st bn knh 1cm mang in tch q nm l lng gia khong khng gian gia hai tm kim loi. Bit khi lng ring ca st l 7800kg/m3 , ca du l 800kg/m3 , ly g = 10m/s2 . Tm du v ln ca q: A. - 12,7 C B. 14,7 C C. - 14,7 C D. 12,7 C Cu hi 5: Mt qu cu khi lng 1g treo u mt si dy mnh cch in. H thng nm trong in trng u c phng nm ngang, cng E = 2kV/m. Khi dy treo hp vi phng thng ng mt gc 600 . Tm in tch ca qu cu, ly g = 10m/s2 : A. 5,8 C B. 6,67 C C. 7,26 C D. 8,67C Cu hi 6: Mt qu cu kim loi nh c khi lng 1g c tch in q = 10-5 C treo vo u mt si dy mnh v t trong in trng u E. Khi qu cu ng cn bng th dy treo hp vi phng thng ng mt gc 600 , ly g = 10m/s2 . Tm E: A. 1730V/m B. 1520V/m C. 1341V/m D. 1124V/m Cu hi 7: Hai qu cu nh mang in tch q1 = - 2nC, q2 = +2nC, c treo u hai si dy cch in di bng nhau trong khng kh ti hai im treo M, N cch nhau 2cm cng mt cao. Khi h cn bng hai dy treo lch khi phng thng ng, mun a cc dy treo v v tr phng thng ng th phi to mt in trng u E c hng no ln bao nhiu: A. Nm ngang hng sang phi, E = 1,5.104 V/m B. Nm ngang hng sang tri, E = 3.104 V/m C. Nm ngang hng sang phi, E = 4,5.104 V/m D. Nm ngang hng sang tri, E = 3,5.104 V/m Cu hi 8: Mt vin bi nh kim loi khi lng 9.10-5 kg th tch 10mm3 c t trong du c khi lng ring 800kg/m3 . Chng t trong in trng u E = 4,1.105 V/m c hng thng ng t trn xung, thy vin bi nm l lng, ly g = 10m/s2 . in tch ca bi l: A. - 1nC B. 1,5nC C. - 2nC D. 2,5nC Cu hi 9: Hai in tch q1 = q2 = q t trong chn khng ln lt ti hai im A v B cch nhau mt khong l. Ti I ngi ta thy in trng ti bng khng. Hi I c v tr no sau y: A. AI = BI = l/2 B. AI = l; BI = 2l C. BI = l; AI = 2l D. AI = l/3; BI = 2l/3 55 M N q1 q2 56. hongthamvp@gmail.com Su tm v bin son Cu hi 10: Hai in tch im q1 = 36 C v q2 = 4 C t trong khng kh ln lt ti hai im A v B cch nhau 100cm. Ti im C in trng tng hp trit tiu, C c v tr no: A. bn trong on AB, cch A 75cm B. bn trong on AB, cch A 60cm C. bn trong on AB, cch A 30cm D. bn trong on AB, cch A 15cm Cu 1 2 3 4 5 6 7 8 9 10 p n C B D C D A C C A A in trng - Dng 3: q cn bng trong in trng, E trit tiu - 2 Cu hi 1: Ba in tch q1, q2, q3 t trong khng kh ln lt ti cc nh A, B, C ca hnh vung ABCD. Bit in trng tng hp ti D trit tiu. Quan h gia 3 in tch trn l: A. q1 = q3; q2 = -2 2 q1 B. q1 = - q3; q2 = 2 2 q1 C. q1 = q3; q2 = 2 2 q1 D. q2 = q3 = - 2 2 q1 Cu hi 2: Mt qu cu khi lng 1g treo u mt si dy mnh cch in. H thng nm trong in trng u c phng nm ngang, cng E = 2kV/m. Khi dy treo hp vi phng thng ng mt gc 600 . Tm sc cng ca si dy, ly g = 10m/s2 : A. 0,01N B. 0,03N C. 0,15N D. 0,02N Cu hi 3: Hai in tch im q v -q t ln lt ti A v B. in trng tng hp trit tiu ti: A. Mt im trong khong AB B. Mt im ngoi khong AB, gn A hn C. Mt im ngoi khong AB, gn B hn D. in trng tng hp khng th trit tiu ti bt c im no Cu hi 4: Hai in tch im q1 v q2 t hai nh A v B ca tam gic u ABC. in trng C bng khng, ta c th kt lun: A. q1 = - q2 B. q1 = q2 C. q1 q2 D. Phi c thm in tch q3 nm u Cu hi 5: Hai in tch im q1 = - q2 = 3C t ln lt ti A v B cch nhau 20cm. in trng tng hp ti trung im O ca AB c: A. ln bng khng B. Hng t O n B, E = 2,7.106 V/m C. Hng t O n A, E = 5,4.106 V/m D. Hng t O n B, E = 5,4.106 V/m Cu hi 6: Hai in tch im q1 = - 2,5 C v q2 = + 6 C t ln lt ti A v B cch nhau 100cm. in trng tng hp trit tiu ti: A. trung im ca AB B. im M trn ng thng AB, ngoi on AB, cch B mt on 1,8m C. im M trn ng thng AB, ngoi on AB, cch A mt on 1,8m D. in trng tng hp khng th trit tiu Cu hi 7: Cc in tch q1 v q2 = q1 t ln lt ti hai nh A v C ca mt hnh vung ABCD. in trng tng hp ti nh D bng khng th phi t ti nh B mt in tch q3 c ln v du bng: A. - q1 B. - 2 q1 C. -2 2 q1 D. khng th tm c v khng bit chiu di ca cnh hnh vung Cu hi 8: Ba in tch im bng nhau q > 0 t ti ba nh ca mt tam gic u ABC. in trng tng hp trit tiu ti: A. mt nh ca tam gic B. tm ca tam gic C. trung im mt cnh ca tam gic D. khng th trit tiu Cu hi 9: Ba in tch im bng nhau q < 0 t ti ba nh ca mt tam gic u ABC. in trng tng hp trit tiu ti: A. mt nh ca tam gic B. tm ca tam gic C. trung im mt cnh ca tam gic D. khng th trit tiu Cu hi 10: Ba in tch im q1, q2 = - 12,5.10-8 C, q3 t ln lt ti A, B, C ca hnh ch nht ABCD cnh AD = a = 3cm, AB = b = 4cm. in trng tng hp ti nh D bng khng. Tnh q1 v q3: A. q1 = 2,7.10-8 C; q3 = 6,4.10-8 C B. q1 = - 2,7.10-8 C; q3 = - 6,4.10-8 C C. q1 = 5,7.10-8 C; q3 = 3,4.10-8 C D. q1 = - 5,7.10-8 C; q3 = - 3,4.10-8 C 56 57. hongthamvp@gmail.com Su tm v bin son Cu 1 2 3 4 5 6 7 8 9 10 p n A D D D D C C B B A A, U, V - Dng 1: Tnh A, U, V ca lc in trng - 1 Cu hi 1: Mt in trng u cng 4000V/m, c phng song song vi cnh huyn BC ca mt tam gic vung ABC c chiu t B n C, bit AB = 6cm, AC = 8cm. Tnh hiu in th gia hai im BC: A. 400V B. 300V C. 200V D. 100V Cu hi 2: Mt in tch q chuyn ng t im M n Q, n N, n P trong in trng u nh hnh v. p n no l sai khi ni v mi quan h gia cng ca lc in trng dch chuyn in tch trn cc on ng: A. AMQ = - AQN B. AMN = ANP C. AQP = AQN D. AMQ = AMP Cu hi 3: Hai tm kim loi phng song song cch nhau 2cm nhim in tri du. Mun lm cho in tch q = 5.10-10 C di chuyn t tm ny sang tm kia cn tn mt cng A = 2.10-9 J. Xc nh cng in trng bn trong hai tm kim loi, bit in trng bn trong l in trng u c ng sc vung gc vi cc tm, khng i theo thi gian: A. 100V/m B. 200V/m C. 300V/m D. 400V/m Cu hi 4: Hiu in th gia hai im M, N l UMN = 2V. Mt in tch q = -1C di chuyn t M n N th cng ca lc in trng l: A. -2J B. 2J C. - 0,5J D. 0,5J Cu hi 5: Mt ht bi khi lng 3,6.10-15 kg mang in tch q = 4,8.10-18 C nm l lng gia hai tm kim loi phng song song nm ngang cch nhau 2cm v nhim in tri du . Ly g = 10m/s2 , tnh hiu in th gia hai tm kim loi: A. 25V. B. 50V C. 75V D. 100V Cu hi 6: Mt qu cu kim loi khi lng 4,5.10-3 kg treo vo u mt si dy di 1m, qu cu nm gia hai tm kim loi phng song song thng ng cch nhau 4cm, t hiu in th gia hai tm l 750V, th qu cu lch 1cm ra khi v tr ban u, ly g = 10m/s2 . Tnh in tch ca qu cu: A. 24nC B. - 24nC C. 48nC D. - 36nC Cu hi 7: Gi thit rng mt tia st c in tch q = 25C c phng t m my dng xung mt t, khi hiu in th gia m my v mt t U = 1,4.108 V. Tnh nng lng ca tia st : A. 35.108 J B. 45.108 J C. 55.108 J D. 65.108 J Cu hi 8: Mt in tch im q = + 10C chuyn ng t nh B n nh C ca tam gic u ABC, nm trong in trng u c cng 5000V/m c ng sc in trng song song vi cnh BC c chiu t C n B. Bit cnh tam gic bng 10cm, tm cng ca lc in trng khi di chuyn in tch trn theo on thng B n C: A. 2,5.10-4 J B. - 2,5.10-4 J C. - 5.10-4 J D. 5.10-4 J Cu hi 9: Mt in tch im q = + 10C chuyn ng t nh B n nh C ca tam gic u ABC, nm trong in trng u c cng 5000V/m c ng sc in trng song song vi cnh BC c chiu t C n B. Bit cnh tam gic bng 10cm, tm cng ca lc in trng khi di chuyn in tch trn theo on gp khc BAC: A. - 10.10-4 J B. - 2,5.10-4 J C. - 5.10-4 J D. 10.10-4 J 57 M Q N P 58. hongthamvp@gmail.com Su tm v bin son Cu hi 10: Mt trong ca mng t bo trong c th sng mang in tch m, mt ngoi mang in tch dng. Hiu in th gia hai mt ny bng 0,07V. Mng t bo dy 8nm. Cng in trng trong mng t bo ny l: A. 8,75.106 V/m B. 7,75.106 V/m C. 6,75.106 V/m D. 5,75.106 V/m Cu 1 2 3 4 5 6 7 8 9 10 p n A D B B C B A C C A A, U, V - Dng 1: Tnh A, U, V ca lc in trng - 2 Cu hi 1: Hai tm kim loi phng nm ngang song song cch nhau 5cm. Hiu in th gia hai tm l 50V. Tnh cng in trng v cho bit c im in trng, dng ng sc in trng gia hai tm kim loi: A. in trng bin i, ng sc l ng cong, E = 1200V/m B. in trng bin i tng dn, ng sc l ng trn, E = 800V/m C. in trng u, ng sc l ng thng, E = 1200V/m D. in trng u, ng sc l ng thng, E = 1000V/m Cu hi 2: Hai tm kim loi phng nm ngang song song cch nhau 5cm. Hiu in th gia hai tm l 50V. Mt electron khng vn tc ban u chuyn ng t tm tch in m v tm tch in dng. Hi khi n tm tch in dng th electron nhn c mt nng lng bng bao nhiu: A. 8.10-18 J B. 7.10-18 J C. 6.10-18 J D. 5.10-18 J Cu hi 3: Cng ca lc in trng lm di chuyn mt in tch gia hai im c hiu in th U = 2000V l 1J. Tnh ln in tch : A. 2mC B. 4.10-2 C C. 5mC D. 5.10-4 C Cu hi 4: Gia hai im A v B c hiu in th bng bao nhiu nu mt in tch q = 1C thu c nng lng 2.10-4 J khi i t A n B: A. 100V B. 200V C. 300V D. 500V Cu hi 5: Cho ba bn kim loi phng tch in 1, 2, 3 t song song ln lt nhau cch nhau nhng khong d12 = 5cm, d23 = 8cm, bn 1 v 3 tch in dng, bn 2 tch in m. E12 = 4.104 V/m, E23 = 5.104 V/m, tnh in th V2, V3 ca cc bn 2 v 3 nu ly gc in th bn 1: A. V2 = 2000V; V3 = 4000V B. V2 = - 2000V; V3 = 4000V C. V2 = - 2000V; V3 = 2000V D. V2 = 2000V; V3 = - 2000V Cu hi 6: Mt qu cu kim loi bn knh 10cm. Tnh in th gy bi qu cu ti im A cch tm qu cu 40cm v ti im B trn mt qu cu, bit in tch ca qu cu l.10-9 C: A. VA = 12,5V; VB = 90V B. VA = 18,2V; VB = 36V C. VA = 22,5V; VB = 76V D.VA = 22,5V; VB = 90V Cu hi 7: Mt qu cu kim loi bn knh 10cm. Tnh in th gy bi qu cu ti im A cch tm qu cu 40cm v ti im B trn mt qu cu, bit in tch ca qu cu l - 5.10-8 C: A. VA = - 4500V; VB = 1125V B. VA = - 1125V; VB = - 4500V C. VA = 1125,5V; VB = 2376V D. VA = 922V; VB = - 5490V Cu hi 8: Mt git thy ngn hnh cu bn knh 1mm tch in q = 3,2.10-13 C t trong khng kh. Tnh cng in trng v in th ca git thy ngn trn b mt git thy ngn: A. 2880V/m; 2,88V B. 3200V/m; 2,88V C. 3200V/m; 3,2V D. 2880; 3,45V Cu hi 9: Mt ht bi kim loi tch in m khi lng 10-10 kg l lng trong khong gia hai bn t in phng nm ngang bn tch in dng trn, bn tch in m di. Hiu in th gia hai bn bng 1000V, khong cch gia hai bn l 4,8mm, ly g = 10m/s2 . Tnh s electron d ht bi: A. 20 000 ht B. 25000 ht C. 30 000 ht D. 40 000 ht 58 59. hongthamvp@gmail.com Su tm v bin son Cu hi 10: Mt in trng u E = 300V/m. Tnh cng ca lc in trng trn di chuyn in tch q = 10nC trn qu o ABC vi ABC l tam gic u cnh a = 10cm nh hnh v: A. 4,5.10-7 J B. 3. 10-7 J C. - 1.5. 10-7 J D. 1.5. 10-7 J Cu 1 2 3 4 5 6 7 8 9 10 p n D A D B C D B A C D A, U, V - Dng 1: Tnh A, U, V ca lc in trng - 3: Cu hi 1: Xt 3 im A, B, C 3 nh ca tam gic vung nh hnh v, = 600 , BC = 6cm, UBC = 120V. Cc hiu in th UAC ,UBA c gi tr ln lt: A. 0; 120V B. - 120V; 0 C. 60 3 V; 60V D. - 60 3 V; 60V Cu hi 2: Mt ht bi khi lng 1g mang in tch - 1C nm yn cn bng trong in trng gia hai bn kim loi phng nm ngang tch in tri du c ln bng nhau. Khong cch gia hai bn l 2cm, ly g = 10m/s2 . Tnh hiu in th gia hai bn kim loi phng trn: A. 20V B. 200V C. 2000V D. 20 000V Cu hi 3: Mt prtn mang in tch + 1,6.10-19 C chuyn ng dc theo phng ca ng sc mt in trng u. Khi n i c qung ng 2,5cm th lc in thc hin mt cng l + 1,6.10-20 J. Tnh cng in trng u ny: A. 1V/m B. 2V/m C. 3V/m D. 4V/m Cu hi 4: Gi thit rng mt tia st c in tch q = 25C c phng t m my dng xung mt t, khi hiu in th gia m my v mt t U = 1,4.108 V. Nng lng ca tia st ny c th lm bao nhiu kilgam nc 1000 C bc thnh hi 1000 C, bit nhit ha hi ca nc bng 2,3.106 J/kg A. 1120kg B. 1521kg C. 2172kg D. 2247kg Cu hi 5: Mt in trng u cng 4000V/m, c phng song song vi cnh huyn BC ca mt tam gic vung ABC c chiu t B n C, bit AB = 6cm, AC = 8cm. Tnh hiu in th gia hai im AC: A. 256V B. 180V C. 128V D. 56V Cu hi 6: Mt in trng u cng 4000V/m, c phng song song vi cnh huyn BC ca mt tam gic vung ABC c chiu t B n C, bit AB = 6cm, AC = 8cm. Tnh hiu in th gia hai im BA: A. 144V B. 120V C. 72V D. 44V Cu hi 7: Cng ca lc in trng lm di chuyn mt in tch gia hai im c hiu in th U = 2000 (V) l A = 1 (J). ln ca in tch l A. q = 2.10-4 (C). B. q = 2.10-4 (C). C. q = 5.10-4 (C). D. q = 5.10-4 (C). Cu hi 8: Hai tm kim loi song song, cch nhau 2 (cm) v c nhim in tri du nhau. Mun lm cho in tch q = 5.10-10 (C) di chuyn t tm ny n tm kia cn tn mt cng A = 2.10-9 (J). Coi in trng bn trong khong gia hai tm kim loi l in trng u v c cc ng sc in vung gc vi cc tm. Cng in trng bn trong tm kim loi l: A. E = 2 (V/m). B. E = 40 (V/m). C. E = 200 (V/m). D. E = 400 (V/m). Cu hi 9: Hai tm kim loi phng nm ngang song song cch nhau 5cm. Hiu in th gia hai tm l 50V. Mt electron khng vn tc ban u chuyn ng t tm tch in m v tm tch in dng. Hi khi n tm tch in dng th electron c vn tc bng bao nhiu: A. 4,2.106 m/s B. 3,2.106 m/s C. 2,2.106 m/s D.1,2.106 m/s Cu hi 10: Cho hai bn kim loi phng t song song tch in tri du, th mt lectron khng vn tc ban u vo in trng gia hai bn kim loi trn. B qua tc dng ca trng trng. Qu o ca lectron l: A. ng thng song song vi cc ng sc in. B. ng thng vung gc vi cc ng sc in. C. mt phn ca ng hypebol. 59 A B C E AB C E 60. hongthamvp@gmail.com Su tm v bin son D. mt phn ca ng parabol. Cu 1 2 3 4 5 6 7 8 9 10 p n A B D B A A B D A B A, U, V - Dng 2: Chuyn ng ca q trong in trng - 1: Cu hi 1: Mt electrn chuyn ng dc theo hng ng sc ca mt in trng u c cng 100V/m vi vn tc ban u l 300 km/s . Hi n chuyn ng c qung ng di bao nhiu th vn tc ca n bng khng: A. 2,56cm B. 25,6cm C. 2,56mm D. 2,56m Cu hi 2: Trong n hnh ca my thu hnh, cc electrn c tng tc bi hiu in th 25 000V. Hi khi p vo mn hnh th vn tc ca n bng bao nhiu, b qua vn tc ban u ca n: A. 6,4.107 m/s B. 7,4.107 m/s C. 8,4.107 m/s D. 9,4.107 m/s Cu hi 3: Mt prtn bay theo phng ca mt ng sc in trng. Lc im A n c vn tc 2,5.104 m/s, khi n im B vn tc ca n bng khng. Bit n c khi lng 1,67.10-27 kg v c in tch 1,6.10-19 C. in th ti A l 500V, tm in th ti B: A. 406,7V B. 500V C. 503,3V D. 533V Cu hi 4: Hai tm kim loi phng nm ngang song song cch nhau 5cm. Hiu in th gia hai tm l 50V. Mt electron khng vn tc ban u chuyn ng t tm tch in m v tm tch in dng. Hi khi n tm tch in dng th electron c vn tc bao nhiu: A. 4,2.106 m/s B. 3,2.106 m/s C. 2,2.106 m/s D. 1,2.106 m/s Cu hi 5: Trong Vt l ht nhn ngi ta hay dng n v nng lng l eV. eV l nng lng m mt electrn thu c khi n i qua on ng c hiu in th 1V. Tnh eV ra Jun, v vn tc ca electrn c nng lng 0,1MeV: A. 1eV = 1,6.1019 J B. 1eV = 22,4.1024 J; C. 1eV = 9,1.10-31 J D. 1eV = 1,6.10-19 J Cu hi 6: Hai bn kim loi phng nm ngang song song cch nhau 10cm c hiu in th gia hai bn l 100V. Mt electrn c vn tc ban u 5.106 m/s chuyn ng dc theo ng sc v bn m. Tnh gia tc ca n. Bit in trng gia hai bn l in trng u v b qua tc dng ca trng lc: A. -17,6.1013 m/s2 B. 15.9.1013 m/s2 C. - 27,6.1013 m/s2 D. + 15,2.1013 m/s2 Cu hi 7: Mt ht bi kim loi tch in m khi lng 10-10 kg l lng trong khong gia hai bn t in phng nm ngang bn tch in dng trn, bn tch in m di. Hiu in th gia hai bn bng 1000V, khong cch gia hai bn l 4,8mm, ly g = 10m/s2 . Chiu tia t ngoi lm ht bi mt mt s electrn v ri xung vi gia tc 6m/s2 . Tnh s ht electrn m ht bi mt: A. 18 000 ht B. 20000 ht C. 24 000 ht D. 28 000 ht Cu hi 8: Mt electrn chuyn ng dc theo mt ng sc ca in trng u c cng 364V/m. Electrn xut pht t im M vi vn tc 3,2.106 m/s i c qung ng di bao nhiu th vn tc ca n bng khng: A. 6cm B. 8cm C. 9cm D. 11cm Cu hi 9: Mt electrn chuyn ng dc theo mt ng sc ca in trng u c cng 364V/m. Electrn xut pht t im M vi vn tc 3,2.106 m/s. Thi gian k t lc xut pht n khi n quay tr v im M l: A. 0,1s B. 0,2 s C. 2 s D. 3 s Cu hi 10: Hai bn kim loi phng nm ngang song song cch nhau 10cm c hiu in th gia hai bn l 100V. Mt electrn c vn tc ban u 5.106 m/s chuyn ng dc theo ng sc v bn m. Tnh on ng n i c cho n khi dng li. Bit in trng gia hai bn l in trng u v b qua tc dng ca trng lc: A. 7,1cm B. 12,2cm C. 5,1cm D. 15,2cm 60 61. hongthamvp@gmail.com Su tm v bin son Cu 1 2 3 4 5 6 7 8 9 10 p n C D C A D A D B A A A, U, V - Dng 2: Chuyn ng ca q trong in trng - 2: Cu hi 1: Mt electrn c phng i t O vi vn tc ban u v0 vung gc vi cc ng sc ca mt in trng u cng E. Khi n im B cch O mt on h theo phng ca ng sc vn tc ca n c biu thc: A. Ehe B. Ehev +2 0 C. Ehev 2 0 D. h m Ee v 22 0 + Cu hi 2: Mt electrn c phng i t O vi vn tc ban u v0 dc theo ng sc ca mt in trng u cng E cng hng in trng. Qung ng xa nht m n di chuyn c trong in trng cho ti khi vn tc ca n bng khng c biu thc: A. Ee mv 2 2 0 B. 2 0 2 mv Ee C. 2 2 0Emve D. 2 0 2 Emve Cu hi 3:Electron chuyn ng khng vn tc ban u t A n B trong mt in trng u vi UAB = 45,5V. Ti B vn tc ca n l: A. 106 m/s2 B. 1,5./s2 C. 4.106 m /s2 D. 8.106 m/s2 Cu hi 4:Khi bay t M n N trong in trng u, electron tng tc ng nng tng thm 250eV. Hiu in th UMN bng: A. -250V B. 250V C. - 125V D. 125V Cu hi 5: Mt t in phng c cc bn nm ngang cch nhau khong d, chiu di cc bn l l. Gia hai bn c hiu in th U. Mt electron bay vo in trng ca t t im O gia cch u hai bn vi vn tc 0v song song vi cc bn. ln gia tc ca n trong in trng l: A. d Ue B. md Ue C. 2 0mdv Ule D. 2 0dv Ule Cu hi 6: Mt t in phng c cc bn nm ngang cch nhau khong d, chiu di cc bn l l. Gia hai bn c hiu in th U. Mt electron bay vo in trng ca t t im O gia cch u hai bn vi vn tc 0v song song vi cc bn. lch ca n theo phng vung gc vi cc bn khi ra khi in trng c biu thc: A. d Ue B. md Ue C. 2 0mdv Ule D. 2 0 2 2mdv Ule Cu hi 7: : Mt t in phng c cc bn nm ngang cch nhau khong d, chiu di cc bn l l. Gia hai bn c hiu in th U. Mt electron bay vo in trng ca t t im O gia cch u hai bn vi vn tc 0v song song vi cc bn. Gc lch gia hng vn tc ca n khi va ra khi in trng v so vi 0v c tan c tnh bi biu thc: A. d Ue B. md Ue C. 2 0mdv Ule D. 2 0 2 2mdv Ule Cu hi 8: Mt electron bay vo in trng ca mt t in phng theo phng song song cng hng vi cc ng sc in trng vi vn tc ban u l 8.106 m/s. Hiu in th t phi c gi tr nh nht l bao nhiu electron khng ti c bn i din A. 182V B.91V C. 45,5V D.50V Cu hi 9: Khi mt electron chuyn ng ngc hng vi vect cng in trng th: A. th nng ca n tng, in th ca n gim B. th nng gim, in th tng 61 62. hongthamvp@gmail.com Su tm v bin son C. th nng v in th u gim D. th nng v in th u tng Cu hi 10: Mt electron c tng tc t trng thi ng yn nh hiu in th U = 200V. Vn tc cui m n t c l: A. 2000m/s B. 8,4.106 m/s C. 2.105 m/s D. 2,1.106 m/s Cu 1 2 3 4 5 6 7 8 9 10 p n D A C A B D C A B B A, U, V - Dng 2: Chuyn ng ca q trong in trng - 3: Cu hi 1: Mt prtn v mt mt electron ln lt c tng tc t trng thi ng yn trong cc in trng u c cng in trng bng nhau v i c nhng qung ng bng nhau th: A. C hai c cng ng nng, electron c gia tc ln hn B. C hai c cng ng nng, electron c gia tc nh hn C. prtn c ng nng ln hn. electron c gia tc ln hn D. electron c ng nng ln hn. Electron c gia tc nh hn Cu hi 2: Mt electron th cho chuyn ng khng vn tc ban u trong in trng u gia hai mt ng th V1 = +10V, V2 = -5V. N s chuyn ng : A. V pha mt ng th V1 B. V pha mt ng th V2 C. Ty cng in trng m n c th v V1 hay V2. D. n ng yn Cu hi 3: Mt electrn c phng i t O vi vn tc ban u v0 dc theo ng sc ca mt in trng u cng E ngc hng in trng. Khi n im B cch O mt on h vn tc ca n c biu thc: A. Ehe B. Ehev +2 0 C. Ehev 2 0 D. h m Ee v 22 0 + Cu hi 4: Trong Vt l ht nhn ngi ta hay dng n v nng lng l eV. eV l nng lng m mt electrn thu c khi n i qua on ng c hiu in th 1V. Tnh vn tc ca electrn c nng lng 0,1MeV: A. v = 0,87.108 m/s B. v = 2,14.108 m/s C. v = 2,87.108 m/s D. v = 1,87.108 m/s Cu hi 5: Hiu in th gia hai im bn ngoi v bn trong ca mt mng t bo l - 90mV, b dy ca mng t bo l 10nm, th in trng( gi s l u) gia mng t bo c cng l: A. 9.106 V/m B. 9.1010 V/m C. 1010 V/m D. 106 V/m Cu hi 6: Khi st nh xung mt t th c mt lng in tch - 30C di chuyn t m my xung mt t. Bit hiu in th gia mt t v m my l 2.107 V. Nng lng m tia st ny truyn t m my xung mt t bng: A. 1,5.10-7 J B. 0,67.107 J C. 6.109 J D. 6.108 J Cu hi 7: Chn mt p n sai : A. Khi mt in tch chuyn ng trn mt mt ng th th cng ca lc in bng khng B. Lc in tc dng ln mt in tch q trong mt mt ng th c phng tip tuyn vi mt ng th C. Vct cng in trng ti mi im trong mt ng th c phng vung gc vi mt ng th D. Khi mt in tch di chuyn t mt mt ng th ny sang mt mt ng th khc th cng ca lc in chc chn khc khng Cu hi 8: Khi electron chuyn ng t bn tch in dng v pha bn m trong khong khng gian gia hai bn kim loi phng tch in tri du ln bng nhau th: A. Lc in thc hin cng dng, th nng lc in tng B. Lc in thc hin cng dng, th nng lc in gim C. Lc in thc hin cng m, th nng lc in tng D. Lc in thc hin cng m, th nng lc in gim Cu hi 9: Hai im A v B nm trn cng mt mt ng th. Mt in tch q chuyn ng t A n B th: A. lc in thc hin cng dng nu q > 0, thc hin cng m nu q < 0 B. lc in thc hin cng dng hay m ty vo du ca q v gi tr in th ca A(B) C. phi bit chiu ca lc in mi xc nh c du ca cng lc in trng D. lc in khng thc hin cng Cu hi 10: Mt in tch +1C chuyn ng t bn tch in dng sang bn tch in m t song song i din nhau th lc in thc hin mt cng bng 200J. Hiu in th gia hai bn c ln bng: A. 5.10-3 V. B. 200V C. 1,6.10-19 V D. 2000V 62 63. hongthamvp@gmail.com Su tm v bin son Cu 1 2 3 4 5 6 7 8 9 10 p n A A D D A C B C D B T in - Dng 1: in dung, nng lng in trng - 1 Cu hi 1: Mt t in in dung 5F c tch in n in tch bng 86C. Tnh hiu in th trn hai bn t: A. 17,2V B. 27,2V C.37,2V D. 47,2V Cu hi 2: Mt t in in dung 24nF tch in n hiu in th 450V th c bao nhiu electron mi di chuyn n bn m ca t in: A. 575.1011 electron B. 675.1011 electron C. 775.1011 electron D. 875.1011 electron Cu hi 3: B t in trong chic n chp nh c in dung 750 F c tch in n hiu in th 330V. Xc nh nng lng m n tiu th trong mi ln n le sng: A. 20,8J B. 30,8J C. 40,8J D. 50,8J Cu hi 4: B t in trong chic n chp nh c in dung 750 F c tch in n hiu in th 330V. Mi ln n le sng t in phng in trong thi gian 5ms. Tnh cng sut phng in ca t in: A. 5,17kW B.6 ,17kW C. 8,17kW D. 8,17kW Cu hi 5:Mt t in c in dung 500pF mc vo hai cc ca mt my pht in c hiu in th 220V. Tnh in tch ca t in: A. 0,31C B. 0,21C C.0,11C D.0,01C Cu hi 6: T in phng khng kh c in dung 5nF. Cng in trng ln nht m t c th chu c l 3.105 V/m, khong cch gia hai bn l 2mm. in tch ln nht c th tch cho t l: A. 2 C B. 3 C C. 2,5C D. 4C Cu hi 7: Nng lng in trng trong t in t l vi: A. hiu in th gia hai bn t in B. in tch trn t in C. bnh phng hiu in th hai bn t in D. hiu in th hai bn t v in tch trn t Cu hi 8: Mt t in c in dung 5nF, in trng ln nht m t c th chu c l 3.105 V/m, khong cch gia hai bn l 2mm. Hiu in th ln nht gia hai bn t l: A. 600V B. 400V C. 500V D.800V Cu hi 9: Mt t in c in dung 2000 pF mc vo hai cc ca ngun in hiu in th 5000V. Tnh in tch ca t in: A. 10C B. 20 C C. 30C D. 40C Cu hi 10: Mt t in c in dung 2000 pF mc vo hai cc ca ngun in hiu in th 5000V. Tch in cho t ri ngt khi ngun, tng in dung t ln hai ln th hiu in th ca t khi l: A. 2500V B. 5000V C. 10 000V D. 1250V Cu 1 2 3 4 5 6 7 8 9 10 p n A B C D C B C A A A T in - Dng 1: in dung, nng lng in trng - 2 63 64. hongthamvp@gmail.com Su tm v bin son Cu hi 1: Mt t in c th chu c in trng gii hn l 3.106 V/m, khong cch gia hai bn t l 1mm, in dung l 8,85.10-11 F. Hi hiu in th ti a c th t vo hai bn t l bao nhiu: A. 3000V B. 300V C. 30 000V D.1500V Cu hi 2: Mt t in c th chu c in trng gii hn l 3.106 V/m, khong cch gia hai bn t l 1mm, in dung l 8,85.10-11 F. Hi in tch cc i m t tch c: A. 26,65.10-8 C B. 26,65.10-9 C C. 26,65.10-7 C D. 13.32. 10-8 C Cu hi 3: T in c in dung 2F c khong cch gia hai bn t l 1cm c tch in vi ngun in c hiu in th 24V. Cng in trng gia hai bn t bng: A. 24V/m B. 2400V/m C. 24 000V/m D. 2,4V Cu hi 4: T in c in dung 2F c khong cch gia hai bn t l 1cm c tch in vi ngun in c hiu in th 24V. Ngt t khi ngun v ni hai bn t bng dy dn th nng lng t gii phng ra l: A. 5,76.10-4 J B. 1,152.10-3 J C. 2,304.10-3 J D.4,217.10-3 J Cu hi 5: Mt t in c in dung C, in tch q, hiu in th U. Tng hiu in th hai bn t ln gp i th in tch ca t: A. khng i B. tng gp i C. tng gp bn D. gim mt na Cu hi 6: Mt t in c in dung C, in tch q, hiu in th U. Ngt t khi ngun, gim in dung xung cn mt na th in tch ca t: A. khng i B. tng gp i C. Gim cn mt na D. gim cn mt phn t Cu hi 7: Mt t in c in dung C, in tch q, hiu in th U. Ngt t khi ngun, gim in dung xung cn mt na th hiu in th gia hai bn t: A. khng i B. tng gp i C. Gim cn mt na D. gim cn mt phn t Cu hi 8: Mt t in c in dung C, in tch q, hiu in th U. Ngt t khi ngun, gim in dung xung cn mt na th nng lng ca t: A. khng i B. tng gp i C. Gim cn mt na D. gim cn mt phn t Cu hi 9: Mt t in phng c in mi l khng kh c in dung l 2F, khong cch gia hai bn t l 1mm. T chu c. Bit in trng gii hn i vi khng kh l 3.106 V/m. Hiu in th v in tch cc i ca t l: A. 1500V; 3mC B. 3000V; 6mC C. 6000V/ 9mC D. 4500V; 9mC Cu hi 10: Mt t in phng c in mi l khng kh c in dung l 2F, khong cch gia hai bn t l 1mm. T chu c. Bit in trng gii hn i vi khng kh l 3.106 V/m. Nng lng ti a m t tch tr c l: A. 4,5J B. 9J C. 18J D. 13,5J Cu 1 2 3 4 5 6 7 8 9 10 p n A A B A B A B B B B T in - Dng 1: in dung, nng lng in trng - 3 Cu hi 1: Mt t in c in dung l bao nhiu th tch ly mt nng lng 0,0015J di mt hiu in th 6V: A. 83,3F B. 1833 F C. 833nF D. 833pF Cu hi 2: Nng lng ca t in tn ti: A. trong khong khng gian gia hai bn t B. hai mt ca bn tch in dng C. hai mt ca bn tch in m D. cc in tch tn ti trn hai bn t Cu hi 3: Mt t in in dung 12pF mc vo ngun in mt chiu c hiu in th 4V. Tng hiu in th ny ln bng 12V th in dung ca t in ny s c gi tr: 64 65. hongthamvp@gmail.com Su tm v bin son A.36pF B. 4pF C. 12pF D. cn ph thuc vo in tch ca t Cu hi 4: Mt t in c in dung 20 F mc vo hiu in th ca ngun mt chiu th in tch ca t bng 80C. Bit hai bn t cch nhau 0,8cm. in trng gia hai bn t c ln: A. 10-4 V/m B. 0,16V/m C. 500V/m D. 5V/m Cu hi 5: Khi t t in c in dung 2 F di hiu in th 5000V th cng thc hin tch in cho t in bng: A. 2,5J B. 5J C. 25J D. 50J Cu hi 6: Vi mt t in xc nh c in dung C khng i, tng nng lng in trng tch tr trong t in ln gp 4 ln ta c th lm cch no sau y: A. tng in tch ca t ln 8 ln, gim hiu in th i 2 ln B. tng hiu in th 8 ln v gim in tch t i 2 ln C. tng hiu in th ln 2 ln D. tng in tch ca t ln 4 ln Cu hi 7: Pht biu no sau y l khng ng? A. T in l h hai vt dn t gn nhau nhng khng tip xc vi nhau. Mi vt gi l mt bn t. B. T in phng l t in c hai bn t l hai tm kim loi c kch thc ln t i din vi nhau. C. in dung ca t in l i lng c trng cho kh nng tch in ca t in v c o bng thng s gia in tch ca t v hiu in th gia hai bn t. D. Hiu in th gii hn l hiu in th ln nht t vo hai bn t in m lp in mi ca t in b nh thng. Cu hi 8: Pht biu no sau y l ng? A. Sau khi np in, t in c nng lng, nng lng tn ti di dng ho nng. B. Sau khi np in, t in c nng lng, nng lng tn ti di dng c nng. C. Sau khi np in, t in c nng lng, nng lng tn ti di dng nhit nng. D. Sau khi np, t in c nng lng, nng lng l nng lng ca in trng trong t in. Cu hi 9: Mt t in c in dung C = 6 (F) c mc vo ngun in 100 (V). Sau khi ngt t in khi ngun, do c qu trnh phng in qua lp in mi nn t in mt dn in tch. Nhit lng to ra trong lp in mi k t khi bt u ngt t in khi ngun in n khi t phng ht in l: A. 0,3 (mJ). B. 30 (kJ). C. 30 (mJ). D. 3.104 (J). Cu hi 10: Mt t in phng c in dung C, c mc vo mt ngun in, sau ngt khi ngun in. Ngi ta nhng hon ton t in vo cht in mi c hng s in mi . Khi hiu in th gia hai bn t in A. Khng thay i. B. Tng ln ln. C. Gim i ln. D. Tng ln hoc gim i tu thuc vo lp in mi. Cu 1 2 3 4 5 6 7 8 9 10 p n A A C C C C A D A A T in - Dng 2: T phng - 1 Cu hi 1: Mt t in phng mc vo hai cc ca mt ngun in c hiu in th 500V. Ngt t khi ngun ri tng khong cch ln hai ln. Hiu in th ca t in khi : A. gim hai ln B. tng hai ln C. tng 4 ln D. gim 4 ln 65 66. hongthamvp@gmail.com Su tm v bin son Cu hi 2: Ni hai bn t in phng vi hai cc ca acquy. Nu dch chuyn cc bn xa nhau th trong khi dch chuyn c dng in i qua acquy khng: A. Khng B. lc u c dng in i t cc m sang cc dng ca acquy sau dng in c chiu ngc li C. dng in i t cc m sang cc dng D. dng in i t cc dng sang cc m Cu hi 3: Ni hai bn t in phng vi hai cc ca ngun mt chiu, sau ngt t ra khi ngun ri a vo gia hai bn mt cht in mi c hng s in mi th in dung C v hiu in th gia hai bn t s: A. C tng, U tng B. C tng, U gim C. C gim, U gim D. C gim, U tng Cu hi 4: Ni hai bn t in phng vi hai cc ca ngun mt chiu, sau ngt t ra khi ngun ri a vo gia hai bn mt cht in mi c hng s in mi th nng lng W ca t v cng in trng E gia hai bn t s: A. W tng; E tng B. W tng; E gim C. Wgim; E gim D. Wgim; E tng Cu hi 5: Mt t in phng c in dung 7nF cha y in mi c hng s in mi , din tch mi bn l 15cm2 v khong cch gia hai bn bng 10-5 m. Tnh hng s in mi : A. 3,7 B. 3,9 C. 4,5 D. 5,3 Cu hi 6: Mt t in phng hai bn c dng hnh trn bn knh 2cm t trong khng kh cch nhau 2mm. in dung ca t in l: A. 1,2pF B. 1,8pF C. 0,87pF D. 0,56pF Cu hi 7: Mt t in phng hai bn c dng hnh trn bn knh 2cm t trong khng kh cch nhau 2mm. C th t mt hiu in th ln nht l bao nhiu vo hai bn t , bit in trng nh nht c th nh thng khng kh l 3.106 V/m: A. 3000V B. 6000V C. 9000V D. 10 000V Cu hi 8: Mt t in phng khng kh mc vo ngun in c hiu in th 200V, din tch mi bn l 20cm2 , hai bn cch nhau 4mm. Tnh mt nng lng in trng trong t in: A. 0,11J/m3 B. 0,27J/m3 C. 0,027J/m3 D. 0,011J/m3 Cu hi 9: in dung ca t in phng ph thuc vo: A. hnh dng, kch thc t v bn cht in mi B. kch thc, v tr tng i ca 2 bn v bn cht in mi C. hnh dng, kch thc, v tr tng i ca hai bn t D. hnh dng, kch thc, v tr tng i ca hai bn t v bn cht in mi Cu hi 10: Hai bn t in phng hnh trn bn knh 60cm, khong cch gia hai bn l 2mm, gia hai bn l khng kh. in dung ca t l: A. 5nF B. 0,5nF C. 50nF D. 5F Cu 1 2 3 4 5 6 7 8 9 10 p n B D B C D D B D D A T in - Dng 3: ghp t - 1 Cu hi 1: Ba t in ging nhau cng in dung C ghp song song vi nhau th in dung ca b t l: 66 67. hongthamvp@gmail.com Su tm v bin son A. C B. 2C C. C/3 D. 3C Cu hi 2: Ba t in ging nhau cng in dung C ghp ni tip vi nhau th in dung ca b t l: A. C B. 2C C. C/3 D. 3C Cu hi 3: B ba t in C1 = C2 = C3/2 ghp song song ri ni vo ngun c hiu in th 45V th in tch ca b t l 18.10-4 C. Tnh in dung ca cc t in: A. C1 = C2 = 5F; C3 = 10 F B. C1 = C2 = 8F; C3 = 16 F C. C1 = C2 = 10F; C3 = 20 F D. C1 = C2 = 15F; C3 = 30 F Cu hi 4: Hai t in c in dung C1 = 2 F; C2 = 3 F mc ni tip nhau. Tnh in dung ca b t: A. 1,8 F B. 1,6 F C. 1,4 F D. 1,2 F Cu hi 5: Hai t in c in dung C1 = 2 F; C2 = 3 F mc ni tip nhau. t vo b t hiu in th mt chiu 50V th hiu in th ca cc t l: A. U1 = 30V; U2 = 20V B. U1 = 20V; U2 = 30V C. U1 = 10V; U2 = 40V D. U1 = 250V; U2 = 25V Cu hi 6: Bn t in mc thnh b theo s nh hnh v, C1 = 1F; C2 = C3 = 3 F. Khi ni hai im M, N vi ngun in th C1 c in tch q1 = 6C v c b t c in tch q = 15,6 C. Hiu in th t vo b t l: A. 4V B. 6V C. 8V D. 10V Cu hi 7: Bn t in mc thnh b theo s nh hnh v trn, C1 = 1F; C2 = C3 = 3 F. Khi ni hai im M, N vi ngun in th C1 c in tch q1 = 6C v c b t c in tch q = 15,6 C. in dung C4 l: A. 1 F B. 2 F C. 3 F D. 4 F Cu hi 8: Ba t C1 = 3nF, C2 = 2nF, C3 = 20nF mc nh hnh v. Ni b t vi hiu in th 30V. Tnh in dung ca c b t: A. 2nF B. 3nF C. 4nF D. 5nF Cu hi 9: Ba t C1 = 3nF, C2 = 2nF, C3 = 20nF mc nh hnh v trn. Ni b